{"id":8088,"date":"2026-09-15T08:24:39","date_gmt":"2026-09-15T07:24:39","guid":{"rendered":"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/calculations-with-solutions-and-gases\/concentrations-of-solutions\/"},"modified":"2026-09-20T09:36:11","modified_gmt":"2026-09-20T08:36:11","slug":"concentrations-of-solutions","status":"publish","type":"page","link":"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/calculations-with-solutions-and-gases\/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-9701-T5-CONCENTRATIONS-OF-SOLUTIONS-REVISION-PAGE-2026\n\nPage:\nConcentrations of Solutions\nCambridge International AS &amp; A Level Chemistry\nPaper 1 and Paper 2\nTopic 2: Atoms, Molecules and Stoichiometry\n9701 and 9701\/02\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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{\n      .ols-image-lightbox {\n        padding: 16px;\n      }\n\n      .ols-image-lightbox-inner {\n        width: 100%;\n        max-height: 88vh;\n      }\n\n      .ols-image-lightbox-img {\n        max-height: 88vh;\n        border-radius: 16px;\n      }\n\n      .ols-image-lightbox-close {\n        top: 10px;\n        right: 10px;\n        width: 42px;\n        height: 42px;\n        font-size: 26px;\n      }\n    }\n\n  <\/style>\n\n  <!--\n    OLS protected content block.\n    Page: Concentrations of Solutions.\n    This content is authored for Online Learning System and must not be scraped,\n    cloned, republished, resold, reformatted into derivative notes, or reused in\n    competing resources without written permission.\n  -->\n\n  <div class=\"ols-revision-layout\">\n    <aside class=\"ols-sidebar\">\n  <div class=\"ols-sidebar-header\">\n    <h3>Revision Notes<\/h3>\n    <p>Cambridge International AS Level Chemistry<\/p>\n  <\/div>\n  <div class=\"ols-topic-group\">\n    <h4>Topic 2 Atoms, Molecules and Stoichiometry<\/h4>\n    <ul class=\"ols-topic-list\">\n      <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/\">Topic overview<\/a><\/li>\n      <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/atoms-elements-and-molecules\/\">2.1 \/ 2.2 Atoms, Elements and Molecules<\/a><\/li>\n<li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/ions-and-ionic-formulae\/\">2.3 Ions and Ionic Formulae<\/a><\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/equations-and-reaction-types\/\">2.3 Equations and Reaction Types<\/a>\n        <ul class=\"ols-subtopic-list\">\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/equations-and-reaction-types\/writing-chemical-equations\/\">Writing Chemical Equations<\/a><\/li>\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/equations-and-reaction-types\/typical-reactions-of-acids\/\">Typical Reactions of Acids<\/a><\/li>\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/equations-and-reaction-types\/ionic-and-full-equations\/\">Ionic and Full Equations<\/a><\/li>\n        <\/ul>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/empirical-and-molecular-formulae\/\">2.3 Empirical and Molecular Formulae<\/a>\n        <ul class=\"ols-subtopic-list\">\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/empirical-and-molecular-formulae\/empirical-formulae\/\">Empirical Formulae<\/a><\/li>\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/empirical-and-molecular-formulae\/molecular-formulae\/\">Molecular Formulae<\/a><\/li>\n<li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/hydrated-salts-and-water-of-crystallisation\/\">2.3 Hydrated Salts and Water of Crystallisation<\/a><\/li>\n        <\/ul>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/mole-calculations\/\">2.4 Mole Calculations<\/a>\n        <ul class=\"ols-subtopic-list\">\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/mole-calculations\/comparing-masses-of-substances\/\">Comparing Masses of Substances<\/a><\/li>\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/mole-calculations\/calculations-involving-moles\/\">Calculations involving Moles<\/a><\/li>\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/mole-calculations\/calculations-using-reacting-masses\/\">Calculations using Reacting Masses<\/a><\/li>\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/mole-calculations\/the-yield-of-a-reaction\/\">The Yield of a Reaction<\/a><\/li>\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/mole-calculations\/atom-economy\/\">Atom Economy<\/a><\/li>\n        <\/ul>\n      <\/li>\n      <li>\n        <a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/calculations-with-solutions-and-gases\/\">2.4 Solutions and Gases<\/a>\n        <ul class=\"ols-subtopic-list\">\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/calculations-with-solutions-and-gases\/molar-volume-calculations\/\">Molar Volume Calculations<\/a><\/li>\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/calculations-with-solutions-and-gases\/concentrations-of-solutions\/\">Concentrations of Solutions<\/a><\/li>\n          <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/calculations-with-solutions-and-gases\/titration-calculations\/\">Titration Calculations<\/a><\/li>\n        <\/ul>\n      <\/li>\n    <\/ul>\n  <\/div>\n  <div class=\"ols-topic-group\">\n    <h4>Other Topics<\/h4>\n    <ul class=\"ols-topic-list\">\n      <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-1-atomic-structure\/\">Topic 1 Atomic Structure<\/a><\/li>\n      <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-3-chemical-bonding\/\">Topic 3 Chemical Bonding<\/a><\/li>\n      <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-4-states-of-matter\/\">Topic 4 States of Matter<\/a><\/li>\n      <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-9-the-periodic-table-chemical-periodicity\/\">Topic 9 Chemical Periodicity<\/a><\/li>\n      <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-13-an-introduction-to-as-level-organic-chemistry\/\">Topic 13 Introduction to Organic Chemistry<\/a><\/li>\n      <li><a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-14-hydrocarbons\/\">Topic 14 Hydrocarbons<\/a><\/li>\n    <\/ul>\n  <\/div>\n<\/aside>\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  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href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/\">A Level Chemistry<\/a> \/\n<a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/\">Cambridge International (CIE)<\/a> \/\n<a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/\">Topic 2 Atoms, Molecules and Stoichiometry<\/a> \/\n<a href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/calculations-with-solutions-and-gases\/\">Calculations with Solutions and Gases<\/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 revision guide to concentrations of solutions for Cambridge International AS &amp; A Level Chemistry Topic 2. This page covers the concentration formula, unit and volume conversions, ion concentrations from dissociation, solution calculations from balanced equations and dilution problems with full worked examples.<\/p>\n\n        <div class=\"ols-badges\">\n<div class=\"ols-badge\">AS Level<\/div>\n<div class=\"ols-badge\">Topic 2: Atoms, Molecules and Stoichiometry<\/div>\n<div class=\"ols-badge\">9701 Papers 1 and 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 how much solute is dissolved in a given volume of solution. In chemistry, it is usually expressed as the <strong>amount<\/strong> of solute (in moles) per unit volume (in dm<sup>3<\/sup>).<\/p>\n\n        <div class=\"ols-equation-box\">\n          <strong>Concentration<\/strong> = amount \u00f7 volume\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>Alternative<\/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=\"Alternative\">M (molar) is the same value<\/td>\n              <\/tr>\n              <tr>\n                <td data-label=\"Quantity\">Amount<\/td>\n                <td data-label=\"Symbol\">n<\/td>\n                <td data-label=\"Standard unit\">mol<\/td>\n                <td data-label=\"Alternative\">Use mol directly<\/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=\"Alternative\">Convert cm<sup>3<\/sup> and m<sup>3<\/sup> as 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>Why dm<sup>3<\/sup> matters:<\/strong> the unit mol dm<sup>\u22123<\/sup> means moles per dm<sup>3<\/sup>. If your volume is given in cm<sup>3<\/sup>, you <strong>must<\/strong> convert it before substituting, or your concentration will be 1000 times too large or too small.<\/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 formula 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 formula 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 moles 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 (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 (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: moles 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>1 mol weighs <strong>84 g<\/strong><\/div>\n            <div class=\"ols-calc-cell\"><span>Amount 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 amount \u00f7 volume method.<\/p>\n        <\/div>\n\n        <div class=\"ols-key-box\">\n          <p><strong>Choose the method that suits you:<\/strong> the direct amount \u00f7 volume 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) + aq \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 the concentration formula 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 moles of the known substance<\/h3>\n            <p>For a solution: amount (mol) = concentration (mol dm<sup>\u22123<\/sup>) \u00d7 volume (dm<sup>3<\/sup>). For a solid: amount (mol) = mass \u00f7 M<sub>r<\/sub>.<\/p>\n          <\/div>\n          <div class=\"ols-rule-item\">\n            <h3>Step 2 &#8211; Apply the mole ratio<\/h3>\n            <p>Use the balancing numbers in the equation to convert moles of the known substance into moles of the substance you need.<\/p>\n          <\/div>\n          <div class=\"ols-rule-item\">\n            <h3>Step 3 &#8211; Convert moles 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>Moles 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>Moles 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 formula 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>Moles 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>moles of solute do 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>Moles 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>amount \u00f7 volume<\/strong>, where amount is in mol and volume is in dm<sup>3<\/sup>.<\/li>\n          <li>The unit mol dm<sup>\u22123<\/sup> is the same as M (molar). Both are acceptable.<\/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>moles = concentration \u00d7 volume in dm<sup>3<\/sup><\/strong>; for solids, <strong>moles = mass \u00f7 M<sub>r<\/sub><\/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 moles of the 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>Moles in a solution:<\/strong> moles = concentration \u00d7 volume in dm<sup>3<\/sup>.<\/li>\n          <li><strong>From mass:<\/strong> moles = mass \u00f7 M<sub>r<\/sub>, 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 moles of the 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 moles of solute do 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 (!lightbox) {\n            return;\n          }\n\n          var lightboxImage = lightbox.querySelector(\".ols-image-lightbox-img\");\n          var closeButton = lightbox.querySelector(\".ols-image-lightbox-close\");\n          var clickableImages = page.querySelectorAll(\"img.ols-lightbox-target\");\n\n          function openLightbox(image) {\n            lightboxImage.src = image.currentSrc || image.src;\n            lightboxImage.alt = image.alt || \"Expanded revision image\";\n            lightbox.classList.add(\"is-open\");\n            lightbox.setAttribute(\"aria-hidden\", \"false\");\n\n            if (!document.body.dataset.olsPreviousOverflow) {\n              document.body.dataset.olsPreviousOverflow = document.body.style.overflow || \"default\";\n            }\n\n            document.body.style.overflow = \"hidden\";\n          }\n\n          function closeLightbox() {\n            lightbox.classList.remove(\"is-open\");\n            lightbox.setAttribute(\"aria-hidden\", \"true\");\n            lightboxImage.src = \"\";\n\n            if (document.body.dataset.olsPreviousOverflow) {\n              document.body.style.overflow = document.body.dataset.olsPreviousOverflow === \"default\" ? \"\" : document.body.dataset.olsPreviousOverflow;\n              delete document.body.dataset.olsPreviousOverflow;\n            }\n          }\n\n          clickableImages.forEach(function(image){\n            image.addEventListener(\"click\", function(event){\n              event.preventDefault();\n              event.stopPropagation();\n              openLightbox(image);\n            });\n          });\n\n          closeButton.addEventListener(\"click\", closeLightbox);\n\n          lightbox.addEventListener(\"click\", function(event){\n            if (event.target === lightbox) {\n              closeLightbox();\n            }\n          });\n\n          document.addEventListener(\"keydown\", function(event){\n            if (event.key === \"Escape\" && lightbox.classList.contains(\"is-open\")) {\n              closeLightbox();\n            }\n          });\n        })();\n      <\/script>\n\n      <!-- FAQs -->\n      <section class=\"ols-faq-card\">\n        <h2>FAQs<\/h2>\n        <p>These questions address the most common points of confusion students encounter when working with solution concentrations.<\/p>\n        <div class=\"ols-faq-list\">\n          <div class=\"ols-faq-item\">\n            <h3>What is the difference between mol dm<sup>\u22123<\/sup> and M?<\/h3>\n            <p>There is no difference; they are two ways of writing the same unit. Both mean moles of solute per dm<sup>3<\/sup> of solution. M is read as &#8220;molar&#8221; (so 0.5 M is &#8220;0.5 molar&#8221;), and mol dm<sup>\u22123<\/sup> is the SI-style notation.<\/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 standard unit of concentration 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 the concentration formula.<\/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 moles of solute do not change when a solution is diluted with water. The product of concentration and volume is the number of moles, so if you multiply concentration by volume before and after, the answers must be equal.<\/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 both its volume and its concentration. Once you have moles of the solution, use the mole ratio from the balanced equation to find moles 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\/cie\/topic-2-atoms-molecules-and-stoichiometry\/calculations-with-solutions-and-gases\/molar-volume-calculations\/\">Molar Volume Calculations<\/a>\n          <a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/calculations-with-solutions-and-gases\/\">Calculations with Solutions and Gases<\/a>\n          <a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/empirical-and-molecular-formulae\/empirical-formulae\/\">Empirical Formula<\/a>\n          <a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/empirical-and-molecular-formulae\/\">Formulae<\/a>\n          <a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/equations-and-reaction-types\/\">Equations and Reaction Types<\/a>\n          <a class=\"ols-related-item\" href=\"https:\/\/www.onlinelearningsystem.net\/xyz\/revision-notes\/a-level-chemistry\/cie\/topic-2-atoms-molecules-and-stoichiometry\/\">Topic 2 Overview<\/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. 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