{"id":5286,"date":"2026-06-19T04:58:13","date_gmt":"2026-06-19T03:58:13","guid":{"rendered":"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/concentration-and-titration-calculations\/concentrations-of-solutions\/"},"modified":"2026-09-20T09:36:08","modified_gmt":"2026-09-20T08:36:08","slug":"concentrations-of-solutions","status":"publish","type":"page","link":"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/concentration-and-titration-calculations\/concentrations-of-solutions\/","title":{"rendered":"Concentrations of Solutions"},"content":{"rendered":"\n<!--\n===============================================================================\nONLINE LEARNING SYSTEM COPYRIGHT NOTICE\n\u00a9 Online Learning System. All rights reserved.\n\nThis WordPress revision page HTML, CSS, content sequence, educational wording,\nlayout structure, schema structure and embedded design logic are protected\nintellectual property of Online Learning System.\n\nUnauthorised copying, redistribution, resale, modification, republication,\nscraping, extraction, derivative reuse, automated harvesting or removal of\ncopyright notices is strictly prohibited.\n\nBackend copyright marker:\nOLS-AQA-312-CONCENTRATIONS-OF-SOLUTIONS-REVISION-PAGE-7405-2026\n\nPage:\nConcentrations of Solutions\nAQA A Level Chemistry\nPaper 1 and Paper 2\n3.1.2 Amount of Substance\n7405\/1 and 7405\/2\n\nCopyright enforcement notes:\n- The visible student page is a free OLS revision resource.\n- The backend HTML structure, CSS architecture, card sequencing, schema graph,\n  revision wording, responsive table behaviour and embedded learning pathway are\n  proprietary OLS production assets.\n- Do not remove this notice.\n===============================================================================\n-->\n\n<section class=\"ols-revision-page ols-concentrations-page\" data-owner=\"Online Learning System\" data-copyright=\"\u00a9 Online Learning System. 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class=\"ols-topic-group\">\r\n    <h4>3.1.2 Amount of Substance<\/h4>\r\n\r\n    <ul class=\"ols-topic-list\">\r\n      <li>\r\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/relative-masses\/\">Relative Masses<\/a>\r\n      <\/li>\r\n\r\n      <li>\r\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/the-mole-and-avogadro-constant\/\">The Mole and Avogadro Constant<\/a>\r\n      <\/li>\r\n\r\n      <li>\r\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/empirical-and-molecular-formulae\/\">Empirical and Molecular Formulae<\/a>\r\n\r\n        <ul class=\"ols-subtopic-list\">\r\n          <li>\r\n            <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/empirical-and-molecular-formulae\/empirical-formula\/\">Empirical Formula<\/a>\r\n          <\/li>\r\n          <li>\r\n            <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/empirical-and-molecular-formulae\/molecular-formula\/\">Molecular Formula<\/a>\r\n          <\/li>\r\n        <\/ul>\r\n      <\/li>\r\n\r\n      <li>\r\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/chemical-equations-and-reacting-masses\/\">Chemical Equations and Reacting Masses<\/a>\r\n\r\n        <ul class=\"ols-subtopic-list\">\r\n          <li>\r\n            <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/chemical-equations-and-reacting-masses\/writing-chemical-equations\/\">Writing Chemical Equations<\/a>\r\n          <\/li>\r\n          <li>\r\n            <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/chemical-equations-and-reacting-masses\/ionic-equations\/\">Ionic Equations<\/a>\r\n          <\/li>\r\n          <li>\r\n            <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/chemical-equations-and-reacting-masses\/calculations-using-reacting-masses\/\">Calculations Using Reacting Masses<\/a>\r\n          <\/li>\r\n        <\/ul>\r\n      <\/li>\r\n\r\n      <li>\r\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/concentration-and-titration-calculations\/\">Concentration and Titration Calculations<\/a>\r\n\r\n        <ul class=\"ols-subtopic-list\">\r\n          <li>\r\n            <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/concentration-and-titration-calculations\/concentrations-of-solutions\/\">Concentrations of Solutions<\/a>\r\n          <\/li>\r\n          <li>\r\n            <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/concentration-and-titration-calculations\/titration-calculations\/\">Titration Calculations<\/a>\r\n          <\/li>\r\n        <\/ul>\r\n      <\/li>\r\n\r\n      <li>\r\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/gas-volumes-and-the-ideal-gas-equation\/\">Gas Volumes and the Ideal Gas Equation<\/a>\r\n      <\/li>\r\n\r\n      <li>\r\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/percentage-yield\/\">Percentage Yield<\/a>\r\n      <\/li>\r\n\r\n      <li>\r\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/atom-economy\/\">Atom Economy<\/a>\r\n      <\/li>\r\n    <\/ul>\r\n  <\/div>\r\n\r\n  <div class=\"ols-topic-group\">\r\n    <h4>Adjacent Topics<\/h4>\r\n\r\n    <ul class=\"ols-topic-list\">\r\n      <li>\r\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-1-atomicstructure\/\">3.1.1 Atomic Structure<\/a>\r\n      <\/li>\r\n      <li>\r\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-3-bonding\/\">3.1.3 Bonding<\/a>\r\n      <\/li>\r\n      <li>\r\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-4-energetics\/\">3.1.4 Energetics<\/a>\r\n      <\/li>\r\n      <li>\r\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-2-1-periodicity\/\">3.2.1 Periodicity<\/a>\r\n      <\/li>\r\n    <\/ul>\r\n  <\/div>\r\n<\/aside>\r\n\r\n<script>\r\n(function() {\r\n  function normalisePath(path) {\r\n    return String(path || '')\r\n      .split('?')[0]\r\n      .split('#')[0]\r\n      .replace(\/\\\/+$\/, '')\r\n      .toLowerCase();\r\n  }\r\n\r\n  function highlightActive() {\r\n    var sidebar = document.querySelector('.ols-sidebar');\r\n    if (!sidebar) return false;\r\n\r\n    var currentPath = normalisePath(window.location.pathname);\r\n    var links = 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href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/\">3.1.2 Amount of Substance<\/a> \/\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/concentration-and-titration-calculations\/\">Concentration and Titration Calculations<\/a> \/\n        <span>Concentrations of Solutions<\/span>\n      <\/nav>\n\n      <header class=\"ols-title-card\">\n        <h1>Concentrations of Solutions<\/h1>\n        <p class=\"ols-page-intro\">A focused AQA revision guide to concentrations of solutions. This page covers concentration in mol dm<sup>\u22123<\/sup>, volume conversions, ion concentrations from dissociation, calculations involving solutions and dilution problems with worked examples.<\/p>\n\n        <div class=\"ols-badges\">\n          <div class=\"ols-badge\">Paper 1 and Paper 2<\/div>\n          <div class=\"ols-badge\">AQA<\/div>\n          <div class=\"ols-badge\">3.1.2 Amount of Substance<\/div>\n          <div class=\"ols-badge\">7405\/1 and 7405\/2<\/div>\n        <\/div>\n\n        <div class=\"ols-author\">\n          <img decoding=\"async\" class=\"ols-author-avatar-img\" src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/05\/Author-Profile.jpeg\" alt=\"Dr. Mohammed Al-Fatah\">\n          <div class=\"ols-author-content\">\n            <h2 class=\"ols-author-title\">Written by:<br><span>Dr. Mohammed Al-Fatah<\/span><\/h2>\n            <p class=\"ols-author-description\">Chemistry specialist revision notes for A Level Chemistry.<\/p>\n            <a class=\"ols-linkedin-pill\" href=\"https:\/\/www.linkedin.com\/in\/doctormohammedfatah\/\" target=\"_blank\" rel=\"noopener noreferrer\">\n              <svg class=\"ols-linkedin-icon\" viewBox=\"0 0 24 24\" fill=\"currentColor\" aria-hidden=\"true\"><path d=\"M4.98 3.5C4.98 4.88 3.86 6 2.48 6S0 4.88 0 3.5 1.12 1 2.48 1s2.5 1.12 2.5 2.5zM.5 8h4V24h-4V8zm7 0h3.8v2.2h.1c.5-.9 1.8-2.2 3.9-2.2 4.2 0 5 2.8 5 6.4V24h-4v-7.6c0-1.8 0-4.2-2.6-4.2s-3 2-3 4v7.8h-4V8z\"><\/path><\/svg>\n              View LinkedIn Profile\n            <\/a>\n          <\/div>\n        <\/div>\n      <\/header>\n\n      <!-- Card 1: How to Work With Solution Concentrations -->\n      <section class=\"ols-h5p-card ols-h5p-inline ols-h5p-recap\">\n<span class=\"ols-h5p-kicker\">Before you start<\/span>\n<h2>GCSE Recap: Concentration in g dm\u207b\u00b3<\/h2>\n<p>Before you start, check that you can convert volumes and work out a concentration in grams per dm\u00b3.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-content\" data-content-id=\"586\"><\/div><\/div>\n<\/section>\n<article class=\"ols-note-card soft\">\n        <div class=\"ols-note-title\"><div class=\"ols-note-icon\">1<\/div><h2>How to Work With Solution Concentrations<\/h2><\/div>\n        <p>The <strong>concentration<\/strong> of a solution tells you the amount of solute dissolved in a given volume of solution. In A Level Chemistry, concentration is usually measured as amount of substance, n, per dm<sup>3<\/sup> of solution.<\/p>\n\n        <div class=\"ols-equation-box\">\n          <strong>c = n \u00f7 V<\/strong><br><strong>n = cV<\/strong>\n        <\/div>\n\n        <p>The standard units you need to know are:<\/p>\n\n        <div class=\"ols-table-wrap\">\n          <table class=\"ols-table\">\n            <thead>\n              <tr>\n                <th>Quantity<\/th>\n                <th>Symbol<\/th>\n                <th>Standard unit<\/th>\n                <th>Calculation note<\/th>\n              <\/tr>\n            <\/thead>\n            <tbody>\n              <tr>\n                <td data-label=\"Quantity\">Concentration<\/td>\n                <td data-label=\"Symbol\">c<\/td>\n                <td data-label=\"Standard unit\">mol dm<sup>\u22123<\/sup><\/td>\n                <td data-label=\"Calculation note\">Use with V in dm<sup>3<\/sup><\/td>\n              <\/tr>\n              <tr>\n                <td data-label=\"Quantity\">Amount of substance<\/td>\n                <td data-label=\"Symbol\">n<\/td>\n                <td data-label=\"Standard unit\">mol<\/td>\n                <td data-label=\"Calculation note\">Use as n<\/td>\n              <\/tr>\n              <tr>\n                <td data-label=\"Quantity\">Volume<\/td>\n                <td data-label=\"Symbol\">V<\/td>\n                <td data-label=\"Standard unit\">dm<sup>3<\/sup><\/td>\n                <td data-label=\"Calculation note\">Convert from cm<sup>3<\/sup> if needed<\/td>\n              <\/tr>\n            <\/tbody>\n          <\/table>\n        <\/div>\n\n        <div class=\"ols-definition-box\">\n          <p><strong>Volume conversions you must memorise:<\/strong><\/p>\n          <p>cm<sup>3<\/sup> \u2192 dm<sup>3<\/sup>: divide by 1000<br>\n          cm<sup>3<\/sup> \u2192 m<sup>3<\/sup>: divide by 1 000 000<br>\n          dm<sup>3<\/sup> \u2192 m<sup>3<\/sup>: divide by 1000<\/p>\n        <\/div>\n\n        <div class=\"ols-key-box\">\n          <p><strong>Unit rule:<\/strong> concentration in mol dm<sup>\u22123<\/sup> uses volume in dm<sup>3<\/sup>. Convert cm<sup>3<\/sup> to dm<sup>3<\/sup> before substituting.<\/p>\n        <\/div>\n      <\/article>\n<section class=\"ols-h5p-card ols-h5p-inline\">\n<span class=\"ols-h5p-kicker\">Check your understanding<\/span>\n<h2>Quick Check: Units and Volume Conversions<\/h2>\n<p>Answer five quick questions on converting volumes and using concentration = amount \u00f7 volume.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-319\" class=\"h5p-iframe\" data-content-id=\"319\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Solution Concentrations Quick Choice: Units and Volume Conversions\"><\/iframe><\/div><\/div>\n<\/section>\n\n      <!-- Image 1: Chemistry solution concentration and conversions (intro foundations) -->\n      <div class=\"ols-zoom-card\">\n        <div class=\"ols-zoom-card-image\">\n          <img decoding=\"async\" src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/05\/Chemistry-solution-concentration-and-conversions.webp\" alt=\"Chemistry solution concentration equation with volume unit conversions between cm3, dm3 and m3\" class=\"ols-zoomable-img ols-lightbox-target\" draggable=\"false\">\n        <\/div>\n        <div class=\"ols-zoom-card-caption\">\n          <p>The concentration equation is short to write and easy to misuse; the real test is whether you can convert the volume to dm<sup>3<\/sup> cleanly every time.<\/p>\n        <\/div>\n      <\/div>\n\n      <!-- Card 2: Calculating Concentration from Mass -->\n      <article class=\"ols-note-card\">\n        <div class=\"ols-note-title\"><div class=\"ols-note-icon\">2<\/div><h2>Calculating Concentration From a Known Mass<\/h2><\/div>\n        <p>Most concentration questions give you a mass of solute and a volume of solution. The strategy is always the same: convert the mass to amount of substance first, then divide by the volume in dm<sup>3<\/sup>.<\/p>\n\n        <div class=\"ols-worked-example\">\n          <h3>Example A &#8211; Small-scale solution<\/h3>\n          <p>Calculate the concentration of the solution made by dissolving <strong>5.00 g of Na<sub>2<\/sub>CO<sub>3<\/sub><\/strong> in water and making the solution up to 250 cm<sup>3<\/sup>.<\/p>\n          <p>M<sub>r<\/sub>(Na<sub>2<\/sub>CO<sub>3<\/sub>) = (23.0 \u00d7 2) + 12 + (16 \u00d7 3) = 106<\/p>\n          <div class=\"ols-calc-row\">\n            <div class=\"ols-calc-cell\"><span>Amount of substance (mol)<\/span>5.00 \u00f7 106 = <strong>0.0472 mol<\/strong><\/div>\n            <div class=\"ols-calc-cell\"><span>Volume (dm<sup>3<\/sup>)<\/span>250 \u00f7 1000 = <strong>0.250 dm<sup>3<\/sup><\/strong><\/div>\n            <div class=\"ols-calc-cell\"><span>Concentration<\/span>0.0472 \u00f7 0.250 = <strong>0.189 mol dm<sup>\u22123<\/sup><\/strong><\/div>\n          <\/div>\n        <\/div>\n\n        <div class=\"ols-worked-example\">\n          <h3>Example B &#8211; Large-scale solution<\/h3>\n          <p>Calculate the concentration of the solution made by dissolving <strong>10 kg of Na<sub>2<\/sub>CO<sub>3<\/sub><\/strong> in water and making the solution up to 0.50 m<sup>3<\/sup>.<\/p>\n          <div class=\"ols-calc-row\">\n            <div class=\"ols-calc-cell\"><span>Amount of substance (mol)<\/span>10 000 \u00f7 106 = <strong>94.2 mol<\/strong><\/div>\n            <div class=\"ols-calc-cell\"><span>Volume (dm<sup>3<\/sup>)<\/span>0.50 \u00d7 1000 = <strong>500 dm<sup>3<\/sup><\/strong><\/div>\n            <div class=\"ols-calc-cell\"><span>Concentration<\/span>94.2 \u00f7 500 = <strong>0.19 mol dm<sup>\u22123<\/sup><\/strong><\/div>\n          <\/div>\n        <\/div>\n\n        <div class=\"ols-key-box\">\n          <p><strong>Sense check:<\/strong> the two solutions above have almost identical concentrations because the ratio of solute to volume is roughly the same. The actual scale of the solution does not change the concentration; only the ratio does.<\/p>\n        <\/div>\n      <\/article>\n<article class=\"ols-note-card soft\">\n<div class=\"ols-note-title\"><div class=\"ols-note-icon\">3<\/div><h2>Mass Concentration in g dm<sup>\u22123<\/sup><\/h2><\/div>\n<p>Concentration can also be given as a <strong>mass concentration<\/strong>, which is the mass of solute in grams per dm<sup>3<\/sup> of solution. The unit is g dm<sup>\u22123<\/sup>.<\/p>\n<div class=\"ols-equation-box\">\n<strong>Mass concentration (g dm<sup>\u22123<\/sup>)<\/strong> = mass of solute (g) \u00f7 volume of solution (dm<sup>3<\/sup>)\n<\/div>\n<p>The two types of concentration are linked by M<sub>r<\/sub>, because the mass of 1 mol of solute is M<sub>r<\/sub> in grams.<\/p>\n<div class=\"ols-equation-box\">\n<strong>Concentration in g dm<sup>\u22123<\/sup><\/strong> = concentration in mol dm<sup>\u22123<\/sup> \u00d7 M<sub>r<\/sub>\n<\/div>\n<div class=\"ols-worked-example\">\n<h3>Worked example: the Na<sub>2<\/sub>CO<sub>3<\/sub> solution from Example A<\/h3>\n<p>The solution contains 5.00 g of Na<sub>2<\/sub>CO<sub>3<\/sub> (M<sub>r<\/sub> = 106) in 250 cm<sup>3<\/sup> of solution.<\/p>\n<div class=\"ols-calc-row\">\n<div class=\"ols-calc-cell\"><span>Volume (dm<sup>3<\/sup>)<\/span>250 \u00f7 1000 = <strong>0.250 dm<sup>3<\/sup><\/strong><\/div>\n<div class=\"ols-calc-cell\"><span>Mass concentration<\/span>5.00 \u00f7 0.250 = <strong>20.0 g dm<sup>\u22123<\/sup><\/strong><\/div>\n<div class=\"ols-calc-cell\"><span>Convert to mol dm<sup>\u22123<\/sup><\/span>20.0 \u00f7 106 = <strong>0.189 mol dm<sup>\u22123<\/sup><\/strong><\/div>\n<\/div>\n<p>This matches the answer to Example A, so both routes give the same concentration.<\/p>\n<\/div>\n<div class=\"ols-key-box\">\n<p><strong>Remember:<\/strong> multiply by M<sub>r<\/sub> to go from mol dm<sup>\u22123<\/sup> to g dm<sup>\u22123<\/sup>. Divide by M<sub>r<\/sub> to go from g dm<sup>\u22123<\/sup> to mol dm<sup>\u22123<\/sup>.<\/p>\n<\/div>\n<\/article>\n\n      <!-- Image 2: Understanding solution concentrations infographic -->\n      <div class=\"ols-zoom-card\">\n        <div class=\"ols-zoom-card-image\">\n          <img decoding=\"async\" src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/05\/Understanding-solution-concentrations-infographic.webp\" alt=\"Understanding solution concentrations infographic comparing molar concentration and mass concentration\" class=\"ols-zoomable-img ols-lightbox-target\" draggable=\"false\">\n        <\/div>\n        <div class=\"ols-zoom-card-caption\">\n          <p>Whether the sample is grams in a beaker or kilograms in an industrial tank, the calculation reduces to the same two numbers: amount of solute and volume in dm<sup>3<\/sup>.<\/p>\n        <\/div>\n      <\/div>\n\n      <!-- Card 3: Alternative Scaling Method -->\n      <article class=\"ols-note-card purple\">\n        <div class=\"ols-note-title\"><div class=\"ols-note-icon\">4<\/div><h2>Alternative Method: Scaling to 1 dm<sup>3<\/sup><\/h2><\/div>\n        <p>Some questions are easier to handle by first scaling the data up to a full 1 dm<sup>3<\/sup> (1000 cm<sup>3<\/sup>) of solution, then converting the resulting mass into moles. This is a useful sanity check when the numbers feel awkward.<\/p>\n\n        <div class=\"ols-worked-example\">\n          <h3>Example C &#8211; Scaling NaHCO<sub>3<\/sub> to 1 dm<sup>3<\/sup><\/h3>\n          <p>What is the concentration in mol dm<sup>\u22123<\/sup> of a solution containing <strong>2.10 g of NaHCO<sub>3<\/sub><\/strong> in 250 cm<sup>3<\/sup> of solution? (H = 1, C = 12, O = 16, Na = 23)<\/p>\n          <p>250 cm<sup>3<\/sup> is one quarter of 1000 cm<sup>3<\/sup> (1 dm<sup>3<\/sup>). So a solution with the same concentration in 1000 cm<sup>3<\/sup> would contain four times as much solute.<\/p>\n          <div class=\"ols-calc-row\">\n            <div class=\"ols-calc-cell\"><span>Mass in 1 dm<sup>3<\/sup><\/span>4 \u00d7 2.10 = <strong>8.40 g<\/strong><\/div>\n            <div class=\"ols-calc-cell\"><span>M<sub>r<\/sub>(NaHCO<sub>3<\/sub>)<\/span>molar mass = <strong>84 g<\/strong><\/div>\n            <div class=\"ols-calc-cell\"><span>Amount of substance in 1 dm<sup>3<\/sup><\/span>8.40 \u00f7 84 = <strong>0.100 mol<\/strong><\/div>\n          <\/div>\n          <p>The concentration is therefore <strong>0.100 mol dm<sup>\u22123<\/sup><\/strong>, which matches the standard c = n \u00f7 V method.<\/p>\n        <\/div>\n\n        <div class=\"ols-key-box\">\n          <p><strong>Choose the method that suits you:<\/strong> the direct c = n \u00f7 V method is more reliable under exam pressure, but the scaling-to-1-dm<sup>3<\/sup> method is a useful way to check that your answer makes sense.<\/p>\n        <\/div>\n      <\/article>\n<section class=\"ols-h5p-card ols-h5p-inline\">\n<span class=\"ols-h5p-kicker\">Check your understanding<\/span>\n<h2>Quick Check: Concentration From a Mass<\/h2>\n<p>Work through six calculations on paper, then flip each card to check your answer and working.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-322\" class=\"h5p-iframe\" data-content-id=\"322\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Solution Concentration Flip Cards: From Mass and Volume to mol dm\u207b\u00b3\"><\/iframe><\/div><\/div>\n<\/section>\n\n      <!-- Card 4: Ion Concentrations from Dissociation -->\n      <article class=\"ols-note-card orange\">\n        <div class=\"ols-note-title\"><div class=\"ols-note-icon\">5<\/div><h2>Ion Concentrations From Dissociation<\/h2><\/div>\n        <p>When an ionic compound dissolves in water, it dissociates into its component ions. The concentration of each ion depends on the stoichiometry of the dissociation, not just on the concentration of the original compound.<\/p>\n\n        <div class=\"ols-worked-example\">\n          <h3>Example D &#8211; Ion concentrations in MgCl<sub>2<\/sub>(aq)<\/h3>\n          <p>If <strong>9.53 g (0.1 mol) of magnesium chloride (MgCl<sub>2<\/sub>)<\/strong> is dissolved in water and made up to 1 dm<sup>3<\/sup> of solution, the concentration of magnesium chloride solution would be <strong>0.1 mol dm<sup>\u22123<\/sup><\/strong>.<\/p>\n          <p>However, MgCl<sub>2<\/sub> dissociates fully on dissolving:<\/p>\n          <p><strong>MgCl<sub>2<\/sub>(s) \u2192 Mg<sup>2+<\/sup>(aq) + 2Cl<sup>\u2212<\/sup>(aq)<\/strong><\/p>\n          <p>So 0.1 mol of MgCl<sub>2<\/sub> produces 0.1 mol of Mg<sup>2+<\/sup> ions and 0.2 mol of Cl<sup>\u2212<\/sup> ions.<\/p>\n          <div class=\"ols-calc-row\">\n            <div class=\"ols-calc-cell\"><span>[MgCl<sub>2<\/sub>]<\/span><strong>0.1 mol dm<sup>\u22123<\/sup><\/strong><\/div>\n            <div class=\"ols-calc-cell\"><span>[Mg<sup>2+<\/sup>]<\/span><strong>0.1 mol dm<sup>\u22123<\/sup><\/strong><\/div>\n            <div class=\"ols-calc-cell\"><span>[Cl<sup>\u2212<\/sup>]<\/span><strong>0.2 mol dm<sup>\u22123<\/sup><\/strong><\/div>\n          <\/div>\n        <\/div>\n\n        <div class=\"ols-key-box\">\n          <p><strong>Rule:<\/strong> the concentration of each ion equals the concentration of the dissolved compound multiplied by the number of those ions in one formula unit. Square brackets, such as [Cl<sup>\u2212<\/sup>], are the standard shorthand for &#8220;concentration of&#8230;&#8221;.<\/p>\n        <\/div>\n      <\/article>\n<section class=\"ols-h5p-card ols-h5p-inline\">\n<span class=\"ols-h5p-kicker\">Check your understanding<\/span>\n<h2>Quick Check: Ion Concentrations<\/h2>\n<p>Write each dissociation equation on paper, then type the concentration asked for.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-587\" class=\"h5p-iframe\" data-content-id=\"587\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Ion Concentrations Flashcards: Working from the Dissociation\"><\/iframe><\/div><\/div>\n<\/section>\n\n      <!-- Image 3: Mass concentration and ions dissociating infographic -->\n      <div class=\"ols-zoom-card\">\n        <div class=\"ols-zoom-card-image\">\n          <img decoding=\"async\" src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/05\/Mass-concentration-and-ions-dissociating-infographic.webp\" alt=\"Mass concentration and ions dissociating infographic showing ionic compound separation in water\" class=\"ols-zoomable-img ols-lightbox-target\" draggable=\"false\">\n        <\/div>\n        <div class=\"ols-zoom-card-caption\">\n          <p>One mole of dissolved salt does not always equal one mole of every ion in solution; the dissociation equation is what determines each individual ion concentration.<\/p>\n        <\/div>\n      <\/div>\n\n      <!-- H5P Card ID 319 -->\n      \n\n      <!-- Card 5: Basic Calculations from Equations Involving Solutions -->\n      <article class=\"ols-note-card\">\n        <div class=\"ols-note-title\"><div class=\"ols-note-icon\">6<\/div><h2>Basic Calculations From Equations Involving Solutions<\/h2><\/div>\n        <p>When a balanced equation involves a solution, you often need to combine c = n \u00f7 V with the mole ratio from the equation. The strategy follows three steps:<\/p>\n        <div class=\"ols-rule-list\">\n          <div class=\"ols-rule-item\">\n            <h3>Step 1 &#8211; Find amount of substance of the known substance<\/h3>\n            <p>For a solution: n = cV. For a solid: n = mass \u00f7 molar mass.<\/p>\n          <\/div>\n          <div class=\"ols-rule-item\">\n            <h3>Step 2 &#8211; Apply the mole ratio<\/h3>\n            <p>Use the coefficients in the balanced equation to convert from the known amount of substance to the required amount of substance.<\/p>\n          <\/div>\n          <div class=\"ols-rule-item\">\n            <h3>Step 3 &#8211; Convert amount of substance to the required answer<\/h3>\n            <p>Convert back into mass, volume of solution or concentration depending on what the question asks for.<\/p>\n          <\/div>\n        <\/div>\n\n        <div class=\"ols-worked-example\">\n          <h3>Example E &#8211; Mass of solid reacting with a solution<\/h3>\n          <p>What is the maximum mass of calcium carbonate that will react with <strong>25.0 cm<sup>3<\/sup> of 2.00 mol dm<sup>\u22123<\/sup> hydrochloric acid<\/strong>? (C = 12, O = 16, Ca = 40)<\/p>\n          <p><strong>Balanced equation:<\/strong> CaCO<sub>3<\/sub> + 2HCl \u2192 CaCl<sub>2<\/sub> + H<sub>2<\/sub>O + CO<sub>2<\/sub><\/p>\n          <div class=\"ols-calc-row\">\n            <div class=\"ols-calc-cell\"><span>Amount of HCl<\/span>(25.0 \u00f7 1000) \u00d7 2.00 = <strong>0.0500 mol<\/strong><\/div>\n            <div class=\"ols-calc-cell\"><span>Mole ratio<\/span>1 mol CaCO<sub>3<\/sub> : 2 mol HCl<\/div>\n            <div class=\"ols-calc-cell\"><span>Amount of CaCO<sub>3<\/sub><\/span>0.0500 \u00f7 2 = <strong>0.0250 mol<\/strong><\/div>\n          <\/div>\n          <p>1 mol of CaCO<sub>3<\/sub> weighs 100 g, so 0.0250 mol weighs <strong>0.0250 \u00d7 100 = 2.50 g<\/strong>.<\/p>\n          <p>The maximum mass of calcium carbonate is therefore <strong>2.50 g<\/strong>.<\/p>\n        <\/div>\n\n        <div class=\"ols-key-box\">\n          <p><strong>Exam tip:<\/strong> in any question where one reactant is given as a volume and concentration of solution, always start by finding the moles of that substance. It is almost always the best starting point.<\/p>\n        <\/div>\n      <\/article>\n<section class=\"ols-h5p-card ols-h5p-inline\">\n<span class=\"ols-h5p-kicker\">Check your understanding<\/span>\n<h2>Quick Check: Reactions Involving Solutions<\/h2>\n<p>Use the three-step method on five new reactions, then flip each card to check your working.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-320\" class=\"h5p-iframe\" data-content-id=\"320\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Reactions of Solutions Flip Cards: Moles, Mole Ratio, Then Mass or Concentration\"><\/iframe><\/div><\/div>\n<\/section>\n\n      <!-- Image 4: Basic calculations involving solutions -->\n      <div class=\"ols-zoom-card\">\n        <div class=\"ols-zoom-card-image\">\n          <img decoding=\"async\" src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/05\/Basic-calculations-involving-solutions.webp\" alt=\"Basic calculations involving solutions showing step-by-step worked chemistry calculations with concentration, moles and balanced equations\" class=\"ols-zoomable-img ols-lightbox-target\" draggable=\"false\">\n        <\/div>\n        <div class=\"ols-zoom-card-caption\">\n          <p>Combining concentration with the mole ratio is the workhorse calculation for titration questions and reactant mass questions alike.<\/p>\n        <\/div>\n      <\/div>\n\n      <!-- H5P Card ID 320 -->\n      \n\n      <!-- Card 6: Calculating Volume of Solution -->\n      <article class=\"ols-note-card purple\">\n        <div class=\"ols-note-title\"><div class=\"ols-note-icon\">7<\/div><h2>Calculating the Volume of a Solution<\/h2><\/div>\n        <p>The concentration equation can be rearranged to find any of the three quantities; concentration, amount or volume.<\/p>\n\n        <div class=\"ols-equation-box\">\n          <strong>Volume (dm<sup>3<\/sup>)<\/strong> = amount (mol) \u00f7 concentration (mol dm<sup>\u22123<\/sup>)\n        <\/div>\n\n        <p>Use this form whenever you know how many moles of a solute you need and the concentration you are working with.<\/p>\n\n        <div class=\"ols-worked-example\">\n          <h3>Example F &#8211; Volume needed to deliver a known mass<\/h3>\n          <p>What volume of <strong>0.500 mol dm<sup>\u22123<\/sup> NaOH<\/strong> contains 4.00 g of sodium hydroxide? (Na = 23, O = 16, H = 1, so M<sub>r<\/sub> = 40)<\/p>\n          <div class=\"ols-calc-row\">\n            <div class=\"ols-calc-cell\"><span>Amount needed<\/span>4.00 \u00f7 40 = <strong>0.100 mol<\/strong><\/div>\n            <div class=\"ols-calc-cell\"><span>Volume (dm<sup>3<\/sup>)<\/span>0.100 \u00f7 0.500 = <strong>0.200 dm<sup>3<\/sup><\/strong><\/div>\n            <div class=\"ols-calc-cell\"><span>Volume (cm<sup>3<\/sup>)<\/span>0.200 \u00d7 1000 = <strong>200 cm<sup>3<\/sup><\/strong><\/div>\n          <\/div>\n        <\/div>\n\n        <div class=\"ols-key-box\">\n          <p><strong>Common error:<\/strong> forgetting to convert the final volume to cm<sup>3<\/sup> when the question asks for it. Read the units in the question stem before writing your final answer.<\/p>\n        <\/div>\n      <\/article>\n<section class=\"ols-h5p-card ols-h5p-inline\">\n<span class=\"ols-h5p-kicker\">Check your understanding<\/span>\n<h2>Quick Check: Rearranging the Concentration Formula<\/h2>\n<p>Answer three questions on finding a volume or a mass from a concentration.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-content\" data-content-id=\"325\"><\/div><\/div>\n<\/section>\n\n      <!-- Image 5: Calculating volume from concentration -->\n      <div class=\"ols-zoom-card\">\n        <div class=\"ols-zoom-card-image\">\n          <img decoding=\"async\" src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/05\/Calculating-volume-from-concentration-guide.webp\" alt=\"Calculating volume from concentration guide showing worked example of solution volume calculation\" class=\"ols-zoomable-img ols-lightbox-target\" draggable=\"false\">\n        <\/div>\n        <div class=\"ols-zoom-card-caption\">\n          <p>Rearranging the same single formula is enough to handle nearly every concentration-based question on the paper, provided the units are consistent.<\/p>\n        <\/div>\n      <\/div>\n\n      <!-- Card 7: Dilutions -->\n      <article class=\"ols-note-card orange\">\n        <div class=\"ols-note-title\"><div class=\"ols-note-icon\">8<\/div><h2>Dilution Calculations<\/h2><\/div>\n        <p>When water is added to a solution, the <strong>amount of solute does not change<\/strong>, but the volume increases, so the concentration falls. This is the principle behind every dilution calculation.<\/p>\n\n        <div class=\"ols-equation-box\">\n          <strong>c<sub>1<\/sub>V<sub>1<\/sub> = c<sub>2<\/sub>V<sub>2<\/sub><\/strong>\n        <\/div>\n\n        <p>Here c<sub>1<\/sub> and V<sub>1<\/sub> are the concentration and volume of the original solution, and c<sub>2<\/sub> and V<sub>2<\/sub> are the concentration and volume of the diluted solution. The units must match on both sides, but they can be in any consistent volume unit (cm<sup>3<\/sup> works fine on both sides).<\/p>\n\n        <div class=\"ols-worked-example\">\n          <h3>Example G &#8211; Volume of water to add<\/h3>\n          <p>What volume of water in cm<sup>3<\/sup> must be added to dilute <strong>5.00 cm<sup>3<\/sup> of 1.00 mol dm<sup>\u22123<\/sup> hydrochloric acid<\/strong> so that it has a concentration of 0.050 mol dm<sup>\u22123<\/sup>?<\/p>\n          <div class=\"ols-calc-row\">\n            <div class=\"ols-calc-cell\"><span>Amount of HCl<\/span>1.00 \u00d7 (5.00 \u00f7 1000) = <strong>0.005 mol<\/strong><\/div>\n            <div class=\"ols-calc-cell\"><span>New total volume<\/span>0.005 \u00f7 0.050 = <strong>0.1 dm<sup>3<\/sup> = 100 cm<sup>3<\/sup><\/strong><\/div>\n            <div class=\"ols-calc-cell\"><span>Volume of water added<\/span>100 \u2212 5 = <strong>95 cm<sup>3<\/sup><\/strong><\/div>\n          <\/div>\n        <\/div>\n\n        <div class=\"ols-key-box\">\n          <p><strong>Critical distinction:<\/strong> the question asks for the <strong>volume of water added<\/strong>, not the <strong>final total volume<\/strong>. Always subtract the starting volume of solution from the final volume to find the water added.<\/p>\n        <\/div>\n      <\/article>\n<section class=\"ols-h5p-card ols-h5p-inline\">\n<span class=\"ols-h5p-kicker\">Check your understanding<\/span>\n<h2>Quick Check: Dilutions<\/h2>\n<p>Answer three questions on what changes, and what does not, when a solution is diluted.<\/p>\n<div class=\"ols-h5p-frame\"><div class=\"h5p-content\" data-content-id=\"324\"><\/div><\/div>\n<\/section>\n\n      <!-- Image 6: Dilutions made easy -->\n      <div class=\"ols-zoom-card\">\n        <div class=\"ols-zoom-card-image\">\n          <img decoding=\"async\" src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/05\/Dilutions-made-easy-a-chemistry-guide.webp\" alt=\"Dilutions made easy chemistry guide showing volumetric flask and concentration change in solutions\" class=\"ols-zoomable-img ols-lightbox-target\" draggable=\"false\">\n        <\/div>\n        <div class=\"ols-zoom-card-caption\">\n          <p>Practically, dilutions are performed in a volumetric flask: a measured volume of stock solution is added, then made up with water to the calibration mark.<\/p>\n        <\/div>\n      <\/div>\n\n      <!-- Card 8: Common Exam Points -->\n      <article class=\"ols-note-card soft\">\n        <div class=\"ols-note-title\"><div class=\"ols-note-icon\">9<\/div><h2>Common Exam Points<\/h2><\/div>\n        <ul>\n          <li>Concentration is <strong>c = n \u00f7 V<\/strong>, where n is amount of substance in mol and V is volume in dm<sup>3<\/sup>.<\/li>\n          <li>Use concentration units of <strong>mol dm<sup>\u22123<\/sup><\/strong> unless another unit is specifically requested.<\/li>\n          <li>Always convert cm<sup>3<\/sup> to dm<sup>3<\/sup> by dividing by 1000 before substituting into the formula.<\/li>\n          <li>For solutions, <strong>n = cV<\/strong>, with V in dm<sup>3<\/sup>; for solids, <strong>n = mass \u00f7 molar mass<\/strong>. Identify which one applies to each substance in the question.<\/li>\n          <li>Ion concentrations depend on the dissociation. For example, in 0.1 mol dm<sup>\u22123<\/sup> MgCl<sub>2<\/sub>, [Mg<sup>2+<\/sup>] = 0.1 mol dm<sup>\u22123<\/sup> and [Cl<sup>\u2212<\/sup>] = 0.2 mol dm<sup>\u22123<\/sup>.<\/li>\n          <li>Square brackets, such as [HCl], mean &#8220;concentration of HCl&#8221; and have units of mol dm<sup>\u22123<\/sup>.<\/li>\n          <li>For reactions involving a solution and a solid, always start by finding the amount of substance you have the most information about.<\/li>\n          <li>For dilutions, use <strong>c<sub>1<\/sub>V<sub>1<\/sub> = c<sub>2<\/sub>V<sub>2<\/sub><\/strong>. Read carefully whether the question asks for the final volume or the volume of water added.<\/li>\n          <li>Show every unit conversion clearly on your script; markers award method marks even if the final number is wrong.<\/li>\n        <\/ul>\n      <\/article>\n\n      <!-- Check Your Understanding bank: H5P 322, 323, 324, 325 -->\n      \n\n      <!-- QuickSnap Summary -->\n      <section class=\"ols-quicksnap-card\">\n        <h2>QuickSnap<\/h2>\n        <p>This text summary condenses the page into the essential exam ideas.<\/p>\n        <ul class=\"ols-quicksnap-list\">\n          <li><strong>Concentration formula:<\/strong> concentration = amount \u00f7 volume, in mol dm<sup>\u22123<\/sup>.<\/li>\n          <li><strong>Volume conversions:<\/strong> cm<sup>3<\/sup> \u2192 dm<sup>3<\/sup> divide by 1000; dm<sup>3<\/sup> \u2192 m<sup>3<\/sup> divide by 1000; cm<sup>3<\/sup> \u2192 m<sup>3<\/sup> divide by 1 000 000.<\/li>\n          <li><strong>Amount in a solution:<\/strong> n = cV, with V in dm<sup>3<\/sup>.<\/li>\n          <li><strong>From mass:<\/strong> n = mass \u00f7 molar mass, then divide by volume in dm<sup>3<\/sup>.<\/li>\n          <li><strong>Mass concentration:<\/strong> concentration in g dm<sup>\u22123<\/sup> = concentration in mol dm<sup>\u22123<\/sup> \u00d7 M<sub>r<\/sub>.<\/li>\n          <li><strong>Ion concentrations:<\/strong> determined by the dissociation equation; multiply by the number of those ions per formula unit.<\/li>\n          <li><strong>Square brackets:<\/strong> [X] means concentration of X in mol dm<sup>\u22123<\/sup>.<\/li>\n          <li><strong>Reactions with solutions:<\/strong> use the mole ratio from the balanced equation, after finding amount of substance you know the most about.<\/li>\n          <li><strong>Dilutions:<\/strong> c<sub>1<\/sub>V<sub>1<\/sub> = c<sub>2<\/sub>V<sub>2<\/sub>. The amount of solute does not change; only the volume.<\/li>\n          <li><strong>Volume of water added:<\/strong> final total volume minus starting volume of solution.<\/li>\n        <\/ul>\n      <\/section>\n\n      <!-- Lightbox element -->\n      <div class=\"ols-image-lightbox\" id=\"olsImageLightboxConcentrations001\" aria-hidden=\"true\" role=\"dialog\" aria-modal=\"true\" aria-label=\"Expanded revision image\">\n        <div class=\"ols-image-lightbox-inner\">\n          <button class=\"ols-image-lightbox-close\" type=\"button\" aria-label=\"Close enlarged image\">\u00d7<\/button>\n          <img decoding=\"async\" class=\"ols-image-lightbox-img\" src=\"\" alt=\"\">\n        <\/div>\n      <\/div>\n\n      <script>\n        (function(){\n          var page = document.querySelector(\".ols-concentrations-page\");\n          if (!page) {\n            return;\n          }\n\n          var lightbox = page.querySelector(\"#olsImageLightboxConcentrations001\");\n          if 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520px;\r\n      }\r\n    }\r\n\r\n    @media (max-width: 760px) {\r\n      .ols-course-cta-aqa-amount-substance {\r\n        margin: 26px 0 22px;\r\n      }\r\n\r\n      .ols-course-cta-aqa-amount-substance .ols-course-cta-card {\r\n        padding: 20px;\r\n        border-radius: 24px;\r\n      }\r\n\r\n      .ols-course-cta-aqa-amount-substance .ols-course-cta-header-row {\r\n        align-items: stretch;\r\n      }\r\n\r\n      .ols-course-cta-aqa-amount-substance .ols-course-button-top {\r\n        width: 100%;\r\n      }\r\n\r\n      .ols-course-cta-aqa-amount-substance .ols-course-cta-stats,\r\n      .ols-course-cta-aqa-amount-substance .ols-course-cta-features {\r\n        grid-template-columns: 1fr;\r\n      }\r\n\r\n      .ols-course-cta-aqa-amount-substance .ols-course-animation-wrap {\r\n        padding: 0;\r\n        border-radius: 0;\r\n      }\r\n\r\n      .ols-course-cta-aqa-amount-substance .ols-animation-frame-shell {\r\n        height: 420px;\r\n        border-radius: 18px;\r\n      }\r\n\r\n      .ols-course-cta-aqa-amount-substance .ols-animation-play-circle {\r\n        width: 76px;\r\n        height: 76px;\r\n      }\r\n\r\n      .ols-course-cta-aqa-amount-substance .ols-animation-play-icon {\r\n        border-top-width: 14px;\r\n        border-bottom-width: 14px;\r\n        border-left-width: 22px;\r\n      }\r\n\r\n      .ols-course-cta-aqa-amount-substance .ols-animation-pause-button {\r\n        left: 12px;\r\n        bottom: 12px;\r\n        min-width: 84px;\r\n        padding: 9px 14px;\r\n        font-size: 13px;\r\n      }\r\n\r\n      .ols-course-cta-aqa-amount-substance .ols-course-button {\r\n        width: 100%;\r\n      }\r\n    }\r\n  <\/style>\r\n\r\n  <div class=\"ols-course-cta-card\">\r\n\r\n    <div class=\"ols-course-cta-top\">\r\n\r\n      <a\r\n        class=\"ols-course-cta-image-link\"\r\n        href=\"[insert here]\"\r\n        aria-label=\"Open the AQA A Level Chemistry 3.1.2 Amount of Substance course page\">\r\n\r\n        <div class=\"ols-course-cta-image\">\r\n          <img decoding=\"async\"\r\n            src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/06\/3-1-2-Amount-of-Substance.jpg\"\r\n            alt=\"AQA A Level Chemistry 3.1.2 Amount of Substance interactive course banner\"\r\n          >\r\n        <\/div>\r\n\r\n      <\/a>\r\n\r\n      <div class=\"ols-course-cta-content\">\r\n\r\n        <div class=\"ols-course-cta-header-row\">\r\n\r\n          <div class=\"ols-course-cta-kicker\">AQA 7405<\/div>\r\n          <div class=\"ols-course-cta-kicker\">Paper 1 &amp; Paper 2<\/div>\r\n          <div class=\"ols-course-cta-kicker\">3.1.2 Amount of Substance<\/div>\r\n\r\n        <\/div>\r\n\r\n        <h2>\r\n          Master Amount of Substance for AQA A Level Chemistry\r\n        <\/h2>\r\n\r\n        <div class=\"ols-course-cta-heading-button\">\r\n          <a\r\n            class=\"ols-course-button ols-course-button-top\"\r\n            href=\"[insert here]\">\r\n            View Course\r\n          <\/a>\r\n        <\/div>\r\n\r\n      <\/div>\r\n\r\n    <\/div>\r\n\r\n    <div class=\"ols-course-cta-intro-stats\">\r\n\r\n      <p class=\"ols-course-cta-intro\">\r\n        Continue from these free revision notes into the full 3.1.2 Amount of Substance course, covering moles, Avogadro constant, empirical and molecular formulae, reacting masses, concentration, titrations, gas volumes, percentage yield and atom economy with guided video teaching, diagnostic MCQ practice, teacher-marked short-answer questions and a personalised progress report.\r\n      <\/p>\r\n\r\n      <div class=\"ols-course-cta-stats\">\r\n\r\n        <div class=\"ols-course-stat\">\r\n          <span>Guided learning<\/span>\r\n          <strong>Coming soon<\/strong>\r\n        <\/div>\r\n\r\n        <div class=\"ols-course-stat\">\r\n          <span>Video lessons<\/span>\r\n          <strong>Coming soon<\/strong>\r\n        <\/div>\r\n\r\n        <div class=\"ols-course-stat\">\r\n          <span>MCQ practice<\/span>\r\n          <strong>Coming soon<\/strong>\r\n        <\/div>\r\n\r\n        <div class=\"ols-course-stat\">\r\n          <span>SAQ practice<\/span>\r\n          <strong>Coming soon<\/strong>\r\n        <\/div>\r\n\r\n      <\/div>\r\n\r\n    <\/div>\r\n\r\n    <div class=\"ols-course-cta-features\">\r\n\r\n      <div class=\"ols-course-feature\">\r\n        <h3>Guided video teaching<\/h3>\r\n        <p>\r\n          Learn the chemistry and exam technique through structured video lessons with worked examples and walkthroughs.\r\n        <\/p>\r\n      <\/div>\r\n\r\n      <div class=\"ols-course-feature\">\r\n        <h3>Instant MCQ feedback<\/h3>\r\n        <p>\r\n          Auto-marked MCQ quizzes provide immediate diagnostic feedback for every answer choice.\r\n        <\/p>\r\n      <\/div>\r\n\r\n      <div class=\"ols-course-feature\">\r\n        <h3>Teacher-marked SAQs<\/h3>\r\n        <p>\r\n          Submit written exam responses and receive chemistry specialist feedback with improvement guidance.\r\n        <\/p>\r\n      <\/div>\r\n\r\n      <div class=\"ols-course-feature\">\r\n        <h3>Progress tracking<\/h3>\r\n        <p>\r\n          Identify strengths and weaknesses across the full 3.1.2 Amount of Substance specification, including mole calculations, formulae, solution calculations, gas calculations, yield and atom economy.\r\n        <\/p>\r\n      <\/div>\r\n\r\n    <\/div>\r\n\r\n    <div class=\"ols-course-animation-wrap\">\r\n\r\n      <h3 class=\"ols-course-animation-title\">\r\n        See how the course works\r\n      <\/h3>\r\n\r\n      <div class=\"ols-animation-frame-shell\">\r\n        <iframe\r\n          id=\"olsCourseAnimationFrame312\"\r\n          title=\"OLS course preview animation\"\r\n          loading=\"lazy\"\r\n          referrerpolicy=\"no-referrer\">\r\n        <\/iframe>\r\n\r\n        <button\r\n          class=\"ols-animation-start-overlay\"\r\n          id=\"olsCourseAnimationStart312\"\r\n          type=\"button\"\r\n          aria-label=\"Play OLS course preview animation\">\r\n          <span class=\"ols-animation-start-content\">\r\n            <span class=\"ols-animation-play-circle\" aria-hidden=\"true\">\r\n              <span class=\"ols-animation-play-icon\"><\/span>\r\n            <\/span>\r\n            <span class=\"ols-animation-start-text\">\r\n              Play course preview animation\r\n            <\/span>\r\n          <\/span>\r\n        <\/button>\r\n\r\n        <button\r\n          class=\"ols-animation-pause-button\"\r\n          id=\"olsCourseAnimationPause312\"\r\n          type=\"button\"\r\n          aria-label=\"Pause course preview animation\">\r\n          Pause\r\n        <\/button>\r\n      <\/div>\r\n\r\n      <p class=\"ols-course-video-status\">\r\n        Click play to start the course preview animation.\r\n      <\/p>\r\n\r\n    <\/div>\r\n\r\n    <div class=\"ols-course-cta-bottom\">\r\n\r\n      <a\r\n        class=\"ols-course-button\"\r\n        href=\"[insert here]\">\r\n        View Course\r\n      <\/a>\r\n\r\n    <\/div>\r\n\r\n  <\/div>\r\n\r\n  <template id=\"olsCourseAnimationTemplate312\">\r\n<!DOCTYPE html>\r\n<html lang=\"en\">\r\n<head>\r\n<meta charset=\"UTF-8\">\r\n<meta name=\"viewport\" content=\"width=device-width, initial-scale=1.0\">\r\n<title>OLS Amount of Substance Promo Animation<\/title>\r\n\r\n<style>\r\n*{box-sizing:border-box;}\r\n\r\nhtml.ols-animation-paused *,\r\nhtml.ols-animation-paused *::before,\r\nhtml.ols-animation-paused *::after{\r\n  animation-play-state:paused!important;\r\n}\r\n\r\n:root{\r\n  --ols-video-screen-w:1180;\r\n  --ols-video-screen-h:664;\r\n  --ols-video-screen-scale:1;\r\n  --ols-mcq-screen-w:1360;\r\n  --ols-mcq-screen-h:1130;\r\n  --ols-mcq-screen-scale:1;\r\n  --ols-saq-screen-w:1360;\r\n  --ols-saq-screen-h:965;\r\n  --ols-saq-screen-scale:1;\r\n}\r\n\r\nhtml,\r\nbody{\r\n  width:100%;\r\n  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left,rgba(70,127,247,0.14),transparent 34%),\r\n    linear-gradient(135deg,#f8fbff 0%,#e9efff 100%);\r\n  opacity:1;\r\n  visibility:visible;\r\n  pointer-events:none;\r\n}\r\n\r\n.ols-video-title-screen.hide{animation:olsVideoTitleFadeOut 0.85s ease forwards;}\r\n.ols-mcq-intro-screen.hide{animation:olsMcqIntroFadeOut 0.85s ease forwards;}\r\n.ols-saq-intro.hide{animation:olsSaqIntroFadeOut 0.85s ease forwards;}\r\n\r\n@keyframes olsVideoTitleFadeOut{to{opacity:0;visibility:hidden;}}\r\n@keyframes olsMcqIntroFadeOut{to{opacity:0;visibility:hidden;}}\r\n@keyframes olsSaqIntroFadeOut{to{opacity:0;visibility:hidden;}}\r\n\r\n.ols-video-title-wrap,\r\n.ols-mcq-intro-title-wrap,\r\n.ols-saq-intro-title-wrap{\r\n  max-width:1250px;\r\n  padding:0 34px;\r\n  text-align:center;\r\n}\r\n\r\n.ols-video-title,\r\n.ols-mcq-intro-title,\r\n.ols-saq-intro-title{\r\n  margin:0;\r\n  font-family:Poppins,Arial,sans-serif;\r\n  font-size:clamp(52px,8vw,124px);\r\n  font-weight:600;\r\n  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    <div class=\"ols-video-title-wrap\">\r\n          <h1 class=\"ols-video-title\">\r\n            <span id=\"videoTitleText\"><\/span><span id=\"videoTitleCursor\" class=\"ols-title-cursor\"><\/span>\r\n          <\/h1>\r\n        <\/div>\r\n      <\/div>\r\n\r\n      <section id=\"videoScene\" class=\"ols-video-scene\">\r\n        <div class=\"ols-video-scale-wrap\">\r\n          <div class=\"ols-video-card\">\r\n            <video\r\n              id=\"lessonVideo\"\r\n              src=\"https:\/\/www.onlinelearningsystem.net\/xyz\/wp-content\/uploads\/2026\/05\/Virtual-Lesson.mp4\"\r\n              muted\r\n              playsinline\r\n              preload=\"auto\">\r\n            <\/video>\r\n          <\/div>\r\n        <\/div>\r\n      <\/section>\r\n    <\/div>\r\n  <\/section>\r\n\r\n  <section id=\"sceneMcq\" class=\"ols-scene\">\r\n    <div class=\"ols-mcq-stage\">\r\n      <div id=\"mcqIntroScreen\" class=\"ols-mcq-intro-screen\">\r\n        <div class=\"ols-mcq-intro-title-wrap\">\r\n          <h1 class=\"ols-mcq-intro-title\">\r\n            <span id=\"mcqIntroTitleText\"><\/span><span id=\"mcqIntroTitleCursor\" class=\"ols-title-cursor\"><\/span>\r\n          <\/h1>\r\n        <\/div>\r\n      <\/div>\r\n\r\n      <div class=\"ols-mcq-scale-wrap\">\r\n        <main id=\"mcqScreen\" class=\"ols-mcq-screen\">\r\n          <section class=\"mcq-question-area\">\r\n            <p class=\"mcq-question-lead\">A student prepares a sodium hydroxide solution.<\/p>\r\n\r\n            <table class=\"mcq-ionic-table\">\r\n              <thead>\r\n                <tr>\r\n                  <th>Quantity<\/th>\r\n                  <th>Value<\/th>\r\n                <\/tr>\r\n              <\/thead>\r\n              <tbody>\r\n                <tr><td>Concentration of NaOH<\/td><td>0.200 mol dm<sup>\u22123<\/sup><\/td><\/tr>\r\n                <tr><td>Volume used<\/td><td>25.0 cm<sup>3<\/sup><\/td><\/tr>\r\n                <tr><td>Volume in dm<sup>3<\/sup><\/td><td>0.0250 dm<sup>3<\/sup><\/td><\/tr>\r\n              <\/tbody>\r\n            <\/table>\r\n\r\n            <p class=\"mcq-question-main\">What amount of NaOH is present in the 25.0 cm<sup>3<\/sup> sample?<\/p>\r\n\r\n            <div class=\"mcq-options-wrap\">\r\n              <div id=\"mcqWrongOption\" class=\"mcq-option-row\">\r\n                <span class=\"mcq-radio\"><\/span>\r\n                <span class=\"mcq-radio-ripple\"><\/span>\r\n                <span class=\"mcq-option-label\">5.00 mol<\/span>\r\n\r\n                <span id=\"mcqSpecificFeedback\" class=\"mcq-specific-feedback\">\r\n                  <span id=\"mcqCrossIcon\" class=\"mcq-cross-icon\">\u00d7<\/span>\r\n                  <span class=\"mcq-specific-feedback-text\">\r\n                    <span id=\"mcqSpecificText\"><\/span><span id=\"mcqSpecificCursor\" class=\"cursor mcq-specific-cursor\" style=\"display:none;\"><\/span>\r\n                  <\/span>\r\n                <\/span>\r\n              <\/div>\r\n\r\n              <div class=\"mcq-option-row\"><span class=\"mcq-radio\"><\/span><span class=\"mcq-radio-ripple\"><\/span><span class=\"mcq-option-label\">0.500 mol<\/span><\/div>\r\n              <div class=\"mcq-option-row\"><span class=\"mcq-radio\"><\/span><span class=\"mcq-radio-ripple\"><\/span><span class=\"mcq-option-label\">0.00500 mol<\/span><\/div>\r\n              <div class=\"mcq-option-row\"><span class=\"mcq-radio\"><\/span><span class=\"mcq-radio-ripple\"><\/span><span class=\"mcq-option-label\">0.000500 mol<\/span><\/div>\r\n            <\/div>\r\n\r\n            <button id=\"mcqSubmitButton\" class=\"mcq-submit-button\">Submit answer<\/button>\r\n          <\/section>\r\n\r\n          <section id=\"mcqGeneralFeedback\" class=\"mcq-general-feedback\">\r\n            <span id=\"mcqGeneralText\"><\/span><span id=\"mcqGeneralCursor\" class=\"cursor mcq-general-cursor\" style=\"display:none;\"><\/span>\r\n          <\/section>\r\n        <\/main>\r\n      <\/div>\r\n    <\/div>\r\n  <\/section>\r\n\r\n  <section id=\"sceneSaq\" class=\"ols-scene\">\r\n    <div class=\"ols-saq-stage\">\r\n      <div id=\"saqIntro\" class=\"ols-saq-intro\">\r\n        <div class=\"ols-saq-intro-title-wrap\">\r\n          <h1 class=\"ols-saq-intro-title\">\r\n            <span id=\"saqIntroTitleText\"><\/span><span id=\"saqIntroTitleCursor\" class=\"ols-title-cursor\"><\/span>\r\n          <\/h1>\r\n        <\/div>\r\n      <\/div>\r\n\r\n      <div class=\"ols-saq-scale-wrap\">\r\n        <main id=\"saqScreen\" class=\"ols-saq-screen\">\r\n          <section class=\"saq-question-area\">\r\n            <div class=\"saq-question-text\">\r\n              <strong>(ii)<\/strong>&nbsp;&nbsp; Calcium carbonate reacts with hydrochloric acid.<br>\r\n              Calculate the mass of CaCO<sub>3<\/sub> that reacts with 25.0 cm<sup>3<\/sup> of 0.200 mol dm<sup>\u22123<\/sup> HCl.<br>\r\n              CaCO<sub>3<\/sub> + 2HCl \u2192 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Convert cm\u00b3 to dm\u00b3 before multiplying by concentration.\";\r\nconst mcqGeneralFeedbackText=\r\n\"Your answer is incorrect.\\n\\n\"+\r\n\"For solution concentration calculations, always convert the volume into dm\u00b3 before using n = cV.\\n\\n\"+\r\n\"\u2022 Volume = 25.0 cm\u00b3 = 25.0 \u00f7 1000 = 0.0250 dm\u00b3.\\n\"+\r\n\"\u2022 Concentration = 0.200 mol dm\u207b\u00b3.\\n\"+\r\n\"\u2022 Amount = concentration \u00d7 volume.\\n\"+\r\n\"\u2022 n = 0.200 \u00d7 0.0250 = 0.00500 mol.\\n\\n\"+\r\n\"The correct answer is: 0.00500 mol\";\r\n\r\nconst mcqIntroScreen=document.getElementById(\"mcqIntroScreen\");\r\nconst mcqIntroTitleText=document.getElementById(\"mcqIntroTitleText\");\r\nconst mcqIntroTitleCursor=document.getElementById(\"mcqIntroTitleCursor\");\r\nconst mcqScreen=document.getElementById(\"mcqScreen\");\r\nconst mcqWrongOption=document.getElementById(\"mcqWrongOption\");\r\nconst mcqSubmitButton=document.getElementById(\"mcqSubmitButton\");\r\nconst mcqSpecificFeedback=document.getElementById(\"mcqSpecificFeedback\");\r\nconst mcqCrossIcon=document.getElementById(\"mcqCrossIcon\");\r\nconst mcqSpecificText=document.getElementById(\"mcqSpecificText\");\r\nconst mcqSpecificCursor=document.getElementById(\"mcqSpecificCursor\");\r\nconst mcqGeneralFeedback=document.getElementById(\"mcqGeneralFeedback\");\r\nconst mcqGeneralText=document.getElementById(\"mcqGeneralText\");\r\nconst mcqGeneralCursor=document.getElementById(\"mcqGeneralCursor\");\r\n\r\nfunction selectMcqWrongOption(){\r\n  mcqWrongOption.classList.add(\"clicking\");\r\n\r\n  setTimeout(function(){\r\n    mcqWrongOption.classList.add(\"selected\");\r\n  },220);\r\n\r\n  setTimeout(function(){\r\n    mcqWrongOption.classList.remove(\"clicking\");\r\n  },850);\r\n}\r\n\r\nfunction clickMcqSubmit(){\r\n  mcqSubmitButton.classList.add(\"clicked\");\r\n\r\n  setTimeout(function(){\r\n    mcqSubmitButton.classList.remove(\"clicked\");\r\n  },240);\r\n}\r\n\r\nfunction showMcqSpecificFeedback(){\r\n  mcqSpecificFeedback.classList.add(\"show\");\r\n\r\n  setTimeout(function(){\r\n    mcqCrossIcon.classList.add(\"show\");\r\n  },180);\r\n\r\n  setTimeout(function(){\r\n    mcqSpecificCursor.style.display=\"inline-block\";\r\n\r\n    typeWriter(mcqSpecificText,mcqSpecificFeedbackText,12,function(){\r\n      mcqSpecificCursor.style.display=\"none\";\r\n\r\n      setTimeout(function(){\r\n        showMcqGeneralFeedback();\r\n      },650);\r\n    });\r\n  },560);\r\n}\r\n\r\nfunction showMcqGeneralFeedback(){\r\n  mcqGeneralFeedback.classList.add(\"show\");\r\n\r\n  setTimeout(function(){\r\n    mcqGeneralCursor.style.display=\"inline-block\";\r\n\r\n    typeWriter(mcqGeneralText,mcqGeneralFeedbackText,8,function(){\r\n      mcqGeneralCursor.style.display=\"none\";\r\n\r\n      setTimeout(function(){\r\n        switchScene(sceneMcq,sceneSaq,function(){\r\n          startSaqIntro();\r\n        });\r\n      },2600);\r\n    });\r\n  },600);\r\n}\r\n\r\nfunction startMcqAnimation(){\r\n  fitMcqScreen();\r\n  mcqScreen.classList.add(\"show\");\r\n\r\n  setTimeout(function(){\r\n    selectMcqWrongOption();\r\n\r\n    setTimeout(function(){\r\n      clickMcqSubmit();\r\n\r\n      setTimeout(function(){\r\n        showMcqSpecificFeedback();\r\n      },620);\r\n    },1150);\r\n  },1300);\r\n}\r\n\r\nfunction startMcqIntro(){\r\n  fitMcqScreen();\r\n\r\n  setTimeout(function(){\r\n    typeWriter(mcqIntroTitleText,mcqIntroTitle,44,function(){\r\n      setTimeout(function(){\r\n        mcqIntroTitleCursor.style.display=\"none\";\r\n\r\n        setTimeout(function(){\r\n          mcqIntroScreen.classList.add(\"hide\");\r\n\r\n          setTimeout(function(){\r\n            startMcqAnimation();\r\n          },850);\r\n        },3000);\r\n      },250);\r\n    });\r\n  },700);\r\n}\r\n\r\nconst saqIntroTitle=\"Chemistry Specialist\\nMarked Exam Feedback\";\r\nconst saqStudentAnswer=\"Moles of HCl = 0.200 \u00d7 0.0250 = 0.00500 mol. The mass of calcium carbonate is 0.00500 \u00d7 100 = 0.500 g.\";\r\nconst saqTeacherFeedbackText=\r\n\"Comment:\\n\"+\r\n\"Your first step is correct: you converted 25.0 cm\u00b3 to 0.0250 dm\u00b3 and calculated 0.00500 mol of HCl. The error is that you did not use the mole ratio in the balanced equation.\\n\\n\"+\r\n\"The equation shows that 1 mol of CaCO\u2083 reacts with 2 mol of HCl. Therefore the amount of CaCO\u2083 is half the amount of HCl.\\n\\n\"+\r\n\"Correct method:\\n\"+\r\n\"\u2022 n(HCl) = cV = 0.200 \u00d7 0.0250 = 0.00500 mol \u2713\\n\"+\r\n\"\u2022 n(CaCO\u2083) = 0.00500 \u00f7 2 = 0.00250 mol \u2713\\n\"+\r\n\"\u2022 M\u1d63(CaCO\u2083) = 40.1 + 12.0 + (3 \u00d7 16.0) = 100.1 \u2713\\n\"+\r\n\"\u2022 mass = n \u00d7 M\u1d63 = 0.00250 \u00d7 100.1 = 0.250 g \u2713\\n\\n\"+\r\n\"Next time, always use the balanced equation after calculating moles. 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Convert volumes in cm<sup>3<\/sup> to dm<sup>3<\/sup> before using c = n \u00f7 V.<\/p>\n          <\/div>\n          <div class=\"ols-faq-item\">\n            <h3>Why do I have to convert cm<sup>3<\/sup> to dm<sup>3<\/sup>?<\/h3>\n            <p>Because the usual concentration unit is mol dm<sup>\u22123<\/sup>. If you leave the volume in cm<sup>3<\/sup>, your concentration will be 1000 times too small. Divide every cm<sup>3<\/sup> value by 1000 before substituting into c = n \u00f7 V.<\/p>\n          <\/div>\n          <div class=\"ols-faq-item\">\n            <h3>How do I find the concentration of individual ions in a solution?<\/h3>\n            <p>Write the dissociation equation for the dissolved compound. The concentration of each ion equals the concentration of the original compound multiplied by the number of those ions in one formula unit. For example, 0.1 mol dm<sup>\u22123<\/sup> Na<sub>2<\/sub>SO<sub>4<\/sub> gives [Na<sup>+<\/sup>] = 0.2 mol dm<sup>\u22123<\/sup> and [SO<sub>4<\/sub><sup>2\u2212<\/sup>] = 0.1 mol dm<sup>\u22123<\/sup>.<\/p>\n          <\/div>\n          <div class=\"ols-faq-item\">\n            <h3>What does c<sub>1<\/sub>V<sub>1<\/sub> = c<sub>2<\/sub>V<sub>2<\/sub> actually mean?<\/h3>\n            <p>It states that the amount of solute does not change when a solution is diluted with water. The product of concentration and volume gives the amount of substance, so cV has the same value before and after dilution.<\/p>\n          <\/div>\n          <div class=\"ols-faq-item\">\n            <h3>For a reaction involving a solid and a solution, where should I start?<\/h3>\n            <p>Start with the substance for which you can directly calculate moles. Usually, that is the solution, because you are given its volume and concentration. Once you have amount of substance in the solution, use the mole ratio from the balanced equation to find the amount of substance of the solid, then convert to mass.<\/p>\n          <\/div>\n          <div class=\"ols-faq-item\">\n            <h3>Does the volume of a dilution include the original solution?<\/h3>\n            <p>Yes. In c<sub>1<\/sub>V<sub>1<\/sub> = c<sub>2<\/sub>V<sub>2<\/sub>, V<sub>2<\/sub> is the <strong>total final volume<\/strong> of the diluted solution, which includes the original solution plus the added water. The volume of water added is V<sub>2<\/sub> \u2212 V<sub>1<\/sub>.<\/p>\n          <\/div>\n        <\/div>\n      <\/section>\n\n      <!-- Related Topics -->\n      <section class=\"ols-related-card\">\n        <h2>Related Topics<\/h2>\n        <p>Build the surrounding skills needed to work confidently with solution chemistry and titration calculations.<\/p>\n        <div class=\"ols-related-grid\">\n          <a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/concentration-and-titration-calculations\/\">Concentration and Titration Calculations<\/a>\n          <a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/concentration-and-titration-calculations\/titration-calculations\/\">Titration Calculations<\/a>\n          <a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/the-mole-and-avogadro-constant\/\">The Mole and Avogadro Constant<\/a>\n          <a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/chemical-equations-and-reacting-masses\/calculations-using-reacting-masses\/\">Calculations Using Reacting Masses<\/a>\n          <a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/gas-volumes-and-the-ideal-gas-equation\/\">Gas Volumes and the Ideal Gas Equation<\/a>\n          <a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/\">3.1.2 Amount of Substance<\/a>\n        <\/div>\n      <\/section>\n\n      <!-- Copyright attribution -->\n      <section class=\"ols-attribution-card\">\n        <p><strong>Copyright notice:<\/strong> This OLS revision content, including the explanations, layout, diagrams, tables and embedded learning structure, is authored for Online Learning System by Dr. Mohammed Al-Fatah. It may not be copied, reproduced, redistributed or adapted without written permission.<\/p>\n      <\/section>\n\n      <!-- JSON-LD Schema -->\n      <script type=\"application\/ld+json\">\n      {\n        \"@context\": \"https:\/\/schema.org\",\n        \"@graph\": [\n          {\n            \"@type\": \"WebPage\",\n            \"@id\": \"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/concentration-and-titration-calculations\/concentrations-of-solutions\/#webpage\",\n            \"url\": \"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/aqa\/3-1-2-amount-of-substance\/concentration-and-titration-calculations\/concentrations-of-solutions\/\",\n            \"name\": \"Concentrations of Solutions - AQA A Level Chemistry Revision Notes\",\n            \"description\": \"AQA A Level Chemistry revision notes covering concentration in mol dm-3, volume conversions, ion concentrations, solution 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