CH1 Chimie \u2014 Quantit\u00e9 de mati\u00e8re et solutions<\/title>\n<style>\n:root{--blue:#113d72;--blue2:#1763aa;--cyan:#eaf6ff;--red:#c62828;--redbg:#fff2f2;--green:#14734d;--greenbg:#edfdf5;--ink:#142033;--muted:#5f6f85;--paper:#fff;--bg:#f4f8ff;--line:#d8e2ef;--yellow:#fff7d7;--violet:#6d28d9;--orange:#f97316}\n*{box-sizing:border-box}html{scroll-behavior:smooth}body{margin:0;background:linear-gradient(135deg,#f6f9ff,#eef6ff);color:var(--ink);font-family:Inter,Segoe UI,Roboto,Arial,sans-serif;line-height:1.58}.page{max-width:1160px;margin:auto;padding:28px 18px 70px}.hero{background:radial-gradient(circle at top right,#76c4ff55,transparent 33%),linear-gradient(135deg,#08254d,#1765ad);color:#fff;border-radius:30px;padding:34px;box-shadow:0 18px 42px #12365d33;position:relative;overflow:hidden}.hero:after{content:\"n = m \/ M\";position:absolute;right:22px;bottom:2px;font-size:64px;font-weight:950;opacity:.13}.tag{display:inline-block;background:#ffffff22;border:1px solid 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quantit\u00e9 de mati\u00e8re <strong>n<\/strong>, exprim\u00e9e en mole.<\/p>\n<p><span class=\"pill\">mole<\/span><span class=\"pill\">masse molaire<\/span><span class=\"pill\">volume molaire<\/span><span class=\"pill\">solution<\/span><span class=\"pill\">concentration<\/span><span class=\"pill\">teneur massique<\/span><\/p>\n<div class=\"toolbar\">\n<button class=\"btn\" onclick=\"speakCourse()\">\u25b6 Lire le cours<\/button>\n<button class=\"btn secondary\" onclick=\"pauseSpeech()\">\u23f8 Pause \/ reprise<\/button>\n<button class=\"btn stop\" onclick=\"stopSpeech()\">\u25a0 Stop<\/button>\n<\/div>\n<div class=\"voiceNote\">Lecture vocale par le navigateur : fonctionne avec Safari, Chrome, Edge, Firefox selon les voix disponibles.<\/div>\n<\/section>\n\n<section class=\"grid\">\n<div class=\"card span4 toc\">\n<h2>Plan<\/h2>\n<a href=\"#mole\">1. Mole et constante d\u2019Avogadro<\/a>\n<a href=\"#masse-molaire\">2. Masse molaire<\/a>\n<a href=\"#calculs\">3. Solide pur<\/a>\n<a href=\"#liquides\">4. Liquide pur<\/a>\n<a href=\"#solutions\">5. Solutions<\/a>\n<a href=\"#teneur\">6. Teneur massique<\/a>\n<a href=\"#gaz\">7. Gaz<\/a>\n<a href=\"#exercices\">8. Exercices type bac<\/a>\n<a href=\"#carte\">9. Carte mentale<\/a>\n<a href=\"#resume\">10. R\u00e9sum\u00e9 + quiz<\/a>\n<\/div>\n<div class=\"card span8\">\n<h2>Capacit\u00e9s travaill\u00e9es<\/h2>\n<ul>\n<li>D\u00e9terminer une masse molaire \u00e0 partir des masses molaires atomiques.<\/li>\n<li>D\u00e9terminer une quantit\u00e9 de mati\u00e8re \u00e0 partir d\u2019une masse.<\/li>\n<li>Utiliser une masse volumique ou une densit\u00e9 pour passer d\u2019un volume \u00e0 une masse.<\/li>\n<li>Utiliser le volume molaire d\u2019un gaz.<\/li>\n<li>D\u00e9terminer une quantit\u00e9 de mati\u00e8re \u00e0 partir d\u2019une concentration et d\u2019un volume de solution.<\/li>\n<\/ul>\n<div class=\"warn\"><strong>Fil rouge du chapitre :<\/strong> identifier la situation, choisir la bonne formule rouge, convertir les unit\u00e9s, puis souligner le r\u00e9sultat avec son unit\u00e9.<\/div>\n<\/div>\n<\/section>\n\n<div id=\"courseText\">\n<section id=\"mole\" class=\"card\">\n<h2>1. La mole : compter les entit\u00e9s chimiques<\/h2>\n<p>Pour compter les mol\u00e9cules, les atomes ou les ions, on utilise un paquet de r\u00e9f\u00e9rence : <strong>la mole<\/strong>.<\/p>\n<div class=\"greenbox\">\n<p><strong>Rappel :<\/strong> <span class=\"red\">1 mole contient<\/span> <strong>N<sub>A<\/sub> = 6,02 \u00d7 10<sup>23<\/sup> entit\u00e9s\u00b7mol<sup>\u22121<\/sup><\/strong>.<\/p>\n<p>N<sub>A<\/sub> est la <strong>constante d\u2019Avogadro<\/strong>.<\/p>\n<\/div>\n<p><span class=\"formula\">N = n \u00d7 N<sub>A<\/sub><\/span> <span class=\"formula blue\">n = N \/ N<sub>A<\/sub><\/span><\/p>\n<table><tr><th>Grandeur<\/th><th>Signification<\/th><th>Unit\u00e9<\/th><\/tr><tr><td>N<\/td><td>nombre d\u2019entit\u00e9s : atomes, ions, mol\u00e9cules<\/td><td>sans unit\u00e9<\/td><\/tr><tr><td>n<\/td><td>quantit\u00e9 de mati\u00e8re<\/td><td>mol<\/td><\/tr><tr><td>N<sub>A<\/sub><\/td><td>constante d\u2019Avogadro<\/td><td>mol<sup>\u22121<\/sup><\/td><\/tr><\/table>\n<div class=\"exo\"><strong>Exemple corrig\u00e9.<\/strong> Si on pr\u00e9l\u00e8ve n = 1,0 \u00d7 10<sup>2<\/sup> mol de cuivre :<div class=\"correction\">N = n \u00d7 N<sub>A<\/sub> = 1,0 \u00d7 10<sup>2<\/sup> \u00d7 6,02 \u00d7 10<sup>23<\/sup> = <span class=\"res\">6,02 \u00d7 10<sup>25<\/sup> atomes<\/span>.<\/div><\/div>\n<\/section>\n\n<section id=\"masse-molaire\" class=\"card\">\n<h2>2. Masse molaire<\/h2>\n<h3>2.1 Masse molaire atomique<\/h3>\n<p>La <strong>masse molaire atomique<\/strong> d\u2019un \u00e9l\u00e9ment est la masse d\u2019une mole d\u2019atomes de cet \u00e9l\u00e9ment.<\/p>\n<div class=\"redbox\"><p><span class=\"red\">\u00c0 retenir :<\/span> une masse molaire s\u2019exprime en <strong>g\u00b7mol<sup>\u22121<\/sup><\/strong>.<\/p><\/div>\n<p>Exemple : M(O) = 16 g\u00b7mol<sup>\u22121<\/sup>.<\/p>\n<h3>2.2 Masse molaire mol\u00e9culaire<\/h3>\n<p>Pour une mol\u00e9cule, on additionne les masses molaires atomiques de tous les atomes pr\u00e9sents.<\/p>\n<p><span class=\"formula\">M(C<sub>x<\/sub>H<sub>y<\/sub>O<sub>z<\/sub>) = xM(C) + yM(H) + zM(O)<\/span><\/p>\n<div class=\"bluebox\"><strong>Exemple : glucose C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub><\/strong><br>M = 6\u00d712 + 12\u00d71 + 6\u00d716 = <span class=\"res\">180 g\u00b7mol<sup>\u22121<\/sup><\/span>.<\/div>\n<div class=\"bluebox\"><strong>Exemple : eau H<sub>2<\/sub>O<\/strong><br>M = 2\u00d71 + 16 = <span class=\"res\">18 g\u00b7mol<sup>\u22121<\/sup><\/span>.<\/div>\n<h3>2.3 Masse molaire ionique<\/h3>\n<p>Pour un ion, <span class=\"red\">on ne tient pas compte de la charge \u00e9lectrique<\/span> dans le calcul de la masse molaire.<\/p>\n<p>Exemple : M(ClO<sup>\u2212<\/sup>) = M(Cl) + M(O).<\/p>\n<\/section>\n\n<section id=\"calculs\" class=\"card\">\n<h2>3. Calculer une quantit\u00e9 de mati\u00e8re \u00e0 partir d\u2019un solide pur<\/h2>\n<p><span class=\"formula\">n = m \/ M<\/span> <span class=\"formula blue\">m = n \u00d7 M<\/span><\/p>\n<table><tr><th>Grandeur<\/th><th>Unit\u00e9 \u00e0 utiliser<\/th><\/tr><tr><td>n<\/td><td>mol<\/td><\/tr><tr><td>m<\/td><td>g<\/td><\/tr><tr><td>M<\/td><td>g\u00b7mol<sup>\u22121<\/sup><\/td><\/tr><\/table>\n<div class=\"exo\"><strong>Exercice type 1 \u2014 glucose.<\/strong> On pr\u00e9l\u00e8ve m = 2,0 g de glucose C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub>. Calculer n.<div class=\"correction\"><span class=\"eq\">n = m \/ M<\/span>M = 180 g\u00b7mol<sup>\u22121<\/sup><br>n = 2,0 \/ 180 = <span class=\"res\">1,1 \u00d7 10<sup>\u22122<\/sup> mol<\/span>.<\/div><\/div>\n<div class=\"exo\"><strong>Exercice type 2 \u2014 chlorure de sodium.<\/strong> Calculer la masse \u00e0 peser pour pr\u00e9lever n = 2,0\u00d710<sup>\u22121<\/sup> mol de NaCl(s). Donn\u00e9es : M(Na)=23,0 g\u00b7mol<sup>\u22121<\/sup>, M(Cl)=35,5 g\u00b7mol<sup>\u22121<\/sup>.<div class=\"correction\">M(NaCl)=23,0+35,5=58,5 g\u00b7mol<sup>\u22121<\/sup>.<br><span class=\"eq\">m = n \u00d7 M<\/span>m = 2,0\u00d710<sup>\u22121<\/sup> \u00d7 58,5 = <span class=\"res\">12 g<\/span>.<\/div><\/div>\n<\/section>\n\n<section id=\"liquides\" class=\"card\">\n<h2>4. Calculer une quantit\u00e9 de mati\u00e8re \u00e0 partir du volume d\u2019un liquide pur<\/h2>\n<p>Pour un liquide pur, on ne peut pas utiliser directement le volume dans <span class=\"red\">n = m \/ M<\/span>. Il faut d\u2019abord trouver la masse du liquide.<\/p>\n<h3>4.1 Avec la masse volumique \u03c1<\/h3>\n<p><span class=\"formula\">\u03c1 = m \/ V<\/span> donc <span class=\"formula blue\">m = \u03c1 \u00d7 V<\/span><\/p>\n<div class=\"warn\"><strong>Unit\u00e9s coh\u00e9rentes :<\/strong> si \u03c1 est en g\u00b7L<sup>\u22121<\/sup>, alors V doit \u00eatre en L et m sera en g.<\/div>\n<h3>4.2 Avec la densit\u00e9 d<\/h3>\n<p>Pour un liquide, la densit\u00e9 compare sa masse volumique \u00e0 celle de l\u2019eau.<\/p>\n<p><span class=\"formula\">d = \u03c1<sub>liquide<\/sub> \/ \u03c1<sub>eau<\/sub><\/span> avec <strong>\u03c1<sub>eau<\/sub> = 1000 g\u00b7L<sup>\u22121<\/sup><\/strong>.<\/p>\n<p>Donc : <span class=\"formula\">\u03c1<sub>liquide<\/sub> = d \u00d7 1000<\/span><\/p>\n<div class=\"exo\"><strong>Exercice type 3 \u2014 \u00e9thanol.<\/strong> D\u00e9terminer la quantit\u00e9 de mati\u00e8re dans V = 10 mL d\u2019\u00e9thanol pur. Donn\u00e9es : \u03c1 = 803 g\u00b7L<sup>\u22121<\/sup> ; M(\u00e9thanol)=46 g\u00b7mol<sup>\u22121<\/sup>.<div class=\"correction\"><span class=\"red\">Conversion :<\/span> V = 10 mL = 10 \u00d7 10<sup>\u22123<\/sup> L = 1,0\u00d710<sup>\u22122<\/sup> L.<br><span class=\"eq\">m = \u03c1 \u00d7 V<\/span>m = 803 \u00d7 1,0\u00d710<sup>\u22122<\/sup> = 8,03 g.<br><span class=\"eq\">n = m \/ M<\/span>n = 8,03 \/ 46 = <span class=\"res\">1,7\u00d710<sup>\u22121<\/sup> mol<\/span>.<\/div><\/div>\n<div class=\"exo\"><strong>Exercice type 4 \u2014 acide sulfurique.<\/strong> Calculer la quantit\u00e9 de mati\u00e8re dans V = 20 mL d\u2019acide sulfurique H<sub>2<\/sub>SO<sub>4<\/sub> pur de densit\u00e9 d = 1,8. Donn\u00e9e : M(H<sub>2<\/sub>SO<sub>4<\/sub>) = 98 g\u00b7mol<sup>\u22121<\/sup>.<div class=\"correction\"><span class=\"eq\">\u03c1 = d \u00d7 1000<\/span>\u03c1 = 1,8 \u00d7 1000 = 1800 g\u00b7L<sup>\u22121<\/sup>.<br>V = 20 mL = 20\u00d710<sup>\u22123<\/sup> L = 2,0\u00d710<sup>\u22122<\/sup> L.<br><span class=\"eq\">m = \u03c1 \u00d7 V<\/span>m = 1800 \u00d7 2,0\u00d710<sup>\u22122<\/sup> = 36 g.<br><span class=\"eq\">n = m \/ M<\/span>n = 36 \/ 98 = <span class=\"res\">3,7\u00d710<sup>\u22121<\/sup> mol<\/span>.<\/div><\/div>\n<\/section>\n\n<section id=\"solutions\" class=\"card\">\n<h2>5. Solutions : concentration en masse et concentration en quantit\u00e9 de mati\u00e8re<\/h2>\n<p>Une solution contient un <strong>solvant<\/strong> et un ou plusieurs <strong>solut\u00e9s<\/strong>. On utilise deux concentrations importantes.<\/p>\n<table><tr><th>Nom<\/th><th>Formule rouge<\/th><th>Unit\u00e9s<\/th><th>Signification<\/th><\/tr><tr><td>Concentration en masse<\/td><td><span class=\"red\">C<sub>m<\/sub> = m \/ V<\/span><\/td><td>g\u00b7L<sup>\u22121<\/sup><\/td><td>masse de solut\u00e9 par litre de solution<\/td><\/tr><tr><td>Concentration en quantit\u00e9 de mati\u00e8re<\/td><td><span class=\"red\">C = n \/ V<\/span><\/td><td>mol\u00b7L<sup>\u22121<\/sup><\/td><td>quantit\u00e9 de mati\u00e8re de solut\u00e9 par litre de solution<\/td><\/tr><\/table>\n<p><span class=\"formula\">n = C \u00d7 V<\/span> <span class=\"formula blue\">m = C<sub>m<\/sub> \u00d7 V<\/span> <span class=\"formula green\">C = C<sub>m<\/sub> \/ M<\/span><\/p>\n<div class=\"warn\"><strong>Volume en litres obligatoire :<\/strong><br>50 mL = <span class=\"red\">50 \u00d7 10<sup>\u22123<\/sup> L<\/span> = <span class=\"red\">5,0 \u00d7 10<sup>\u22122<\/sup> L<\/span>.<br>On \u00e9crit d\u2019abord \u00ab \u00d710<sup>\u22123<\/sup> \u00bb pour bien retenir que <strong>1 mL = 10<sup>\u22123<\/sup> L<\/strong>.<\/div>\n<div class=\"exo\"><strong>Exercice type 5 \u2014 solution de glucose.<\/strong> Une solution contient 9,0 g de glucose dans V = 250 mL. Calculer C<sub>m<\/sub>, puis C.<div class=\"correction\"><span class=\"red\">Conversion :<\/span> V = 250 mL = 250\u00d710<sup>\u22123<\/sup> L = 0,250 L.<br><span class=\"eq\">C<sub>m<\/sub> = m \/ V<\/span>C<sub>m<\/sub> = 9,0 \/ 0,250 = <span class=\"res\">36 g\u00b7L<sup>\u22121<\/sup><\/span>.<br>M(glucose)=180 g\u00b7mol<sup>\u22121<\/sup>.<br><span class=\"eq\">C = C<sub>m<\/sub> \/ M<\/span>C = 36 \/ 180 = <span class=\"res\">0,20 mol\u00b7L<sup>\u22121<\/sup><\/span>.<\/div><\/div>\n<\/section>\n\n<section id=\"teneur\" class=\"card\">\n<h2>6. Teneur massique, pourcentage massique et vinaigre<\/h2>\n<p>La <strong>teneur massique<\/strong>, ou <strong>pourcentage massique<\/strong>, indique la masse de solut\u00e9 contenue dans une masse donn\u00e9e de solution.<\/p>\n<div class=\"redbox\"><p><span class=\"red\">Formule :<\/span><\/p><p><span class=\"formula\">t = m<sub>solut\u00e9<\/sub> \/ m<sub>solution<\/sub><\/span><\/p><p>Si on l\u2019exprime en pourcentage :<\/p><p><span class=\"formula\">t(%) = (m<sub>solut\u00e9<\/sub> \/ m<sub>solution<\/sub>) \u00d7 100<\/span><\/p><\/div>\n<div class=\"greenbox\"><strong>Interpr\u00e9tation :<\/strong> un vinaigre \u00e0 6,0 % en masse signifie <span class=\"red\">6,0 g d\u2019acide \u00e9thano\u00efque pour 100 g de vinaigre<\/span>.<\/div>\n<div class=\"warn\"><strong>Astuce tr\u00e8s efficace :<\/strong> raisonner sur <strong>1,00 L de solution<\/strong>. On calcule alors la masse d\u2019un litre de vinaigre avec la masse volumique, puis on applique le pourcentage massique.<\/div>\n<p>Donn\u00e9e utilis\u00e9e ici : <strong>\u03c1(vinaigre) = 1,01\u00d710<sup>3<\/sup> g\u00b7L<sup>\u22121<\/sup><\/strong>. Cette valeur est r\u00e9aliste pour un vinaigre alimentaire et permet de faire un calcul plus complet.<\/p>\n\n<div class=\"exo\"><strong>Exercice type \u2014 concentration d\u2019un vinaigre \u00e0 6,0 %.<\/strong><br>Un vinaigre contient 6,0 % en masse d\u2019acide \u00e9thano\u00efque CH<sub>3<\/sub>COOH. Sa masse volumique vaut \u03c1 = 1,01\u00d710<sup>3<\/sup> g\u00b7L<sup>\u22121<\/sup>. Calculer la concentration en quantit\u00e9 de mati\u00e8re C en acide \u00e9thano\u00efque.<div class=\"correction\">\n<h4>M\u00e9thode 1 \u2014 avec calculs interm\u00e9diaires<\/h4>\n<p><span class=\"red\">On raisonne sur V = 1,00 L de vinaigre.<\/span><\/p>\n<span class=\"eq\">m<sub>solution<\/sub> = \u03c1 \u00d7 V<\/span>\n<p>m<sub>solution<\/sub> = 1,01\u00d710<sup>3<\/sup> \u00d7 1,00 = 1,01\u00d710<sup>3<\/sup> g = 1010 g.<\/p>\n<span class=\"eq\">m<sub>solut\u00e9<\/sub> = (t(%) \/ 100) \u00d7 m<sub>solution<\/sub><\/span>\n<p>m<sub>acide<\/sub> = (6,0 \/ 100) \u00d7 1010 = 60,6 g.<\/p>\n<p>M(CH<sub>3<\/sub>COOH) = 2\u00d712 + 4\u00d71 + 2\u00d716 = 60 g\u00b7mol<sup>\u22121<\/sup>.<\/p>\n<span class=\"eq\">n = m \/ M<\/span>\n<p>n = 60,6 \/ 60 = 1,01 mol.<\/p>\n<span class=\"eq\">C = n \/ V<\/span>\n<p>C = 1,01 \/ 1,00 = <span class=\"res\">1,01 mol\u00b7L<sup>\u22121<\/sup><\/span>.<\/p>\n<h4>M\u00e9thode 2 \u2014 r\u00e9solution litt\u00e9rale<\/h4>\n<p><strong>\u00c9quation 1 :<\/strong> <span class=\"red\">m<sub>solution<\/sub> = \u03c1V<\/span><\/p>\n<p><strong>\u00c9quation 2 :<\/strong> <span class=\"red\">m<sub>solut\u00e9<\/sub> = (t\/100) \u00d7 m<sub>solution<\/sub><\/span><\/p>\n<p><strong>\u00c9quation 3 :<\/strong> <span class=\"red\">n = m<sub>solut\u00e9<\/sub> \/ M<\/span><\/p>\n<p><strong>\u00c9quation 4 :<\/strong> <span class=\"red\">C = n \/ V<\/span><\/p>\n<p>En injectant 1 dans 2, puis 2 dans 3, puis 3 dans 4 :<\/p>\n<span class=\"eq\">C = [(t\/100) \u00d7 \u03c1 \u00d7 V] \/ (M \u00d7 V)<\/span>\n<p>Le volume V se simplifie :<\/p>\n<span class=\"eq\">C = (t\/100) \u00d7 \u03c1 \/ M<\/span>\n<p>C = (6,0\/100) \u00d7 1,01\u00d710<sup>3<\/sup> \/ 60 = <span class=\"res\">1,01 mol\u00b7L<sup>\u22121<\/sup><\/span>.<\/p>\n<\/div><\/div>\n<\/section>\n\n<section id=\"gaz\" class=\"card\">\n<h2>7. Quantit\u00e9 de mati\u00e8re d\u2019un gaz : volume molaire<\/h2>\n<p>Le <strong>volume molaire<\/strong> V<sub>m<\/sub> est le volume occup\u00e9 par une mole de gaz dans des conditions donn\u00e9es de temp\u00e9rature et de pression.<\/p>\n<div class=\"redbox\"><p><span class=\"red\">\u00c0 25 \u00b0C et sous pression atmosph\u00e9rique :<\/span> V<sub>m<\/sub> \u2248 <strong>22,4 L\u00b7mol<sup>\u22121<\/sup><\/strong> dans ce cours.<\/p><\/div>\n<p><span class=\"formula\">n = V \/ V<sub>m<\/sub><\/span> <span class=\"formula blue\">V = n \u00d7 V<sub>m<\/sub><\/span><\/p>\n<div class=\"exo\"><strong>Exercice type 6 \u2014 dioxyg\u00e8ne.<\/strong> Une r\u00e9action produit n = 1,0\u00d710<sup>\u22122<\/sup> mol de dioxyg\u00e8ne O<sub>2<\/sub>. Quel volume de gaz est d\u00e9gag\u00e9 ?<div class=\"correction\"><span class=\"eq\">V = n \u00d7 V<sub>m<\/sub><\/span>V = 1,0\u00d710<sup>\u22122<\/sup> \u00d7 22,4 = <span class=\"res\">2,24\u00d710<sup>\u22121<\/sup> L<\/span>, soit <span class=\"res\">224 mL<\/span>.<\/div><\/div>\n<div class=\"exo\"><strong>Exercice type 7 \u2014 r\u00e9servoir.<\/strong> Un r\u00e9servoir contient V = 50 L de gaz \u00e0 pression atmosph\u00e9rique. Calculer la quantit\u00e9 de mati\u00e8re.<div class=\"correction\"><span class=\"eq\">n = V \/ V<sub>m<\/sub><\/span>n = 50 \/ 22,4 = <span class=\"res\">2,2 mol<\/span>.<\/div><\/div>\n<\/section>\n\n<section id=\"exercices\" class=\"card\">\n<h2>8. Exercices type bac \u2014 entra\u00eenement<\/h2>\n<p class=\"small\">Exercices inspir\u00e9s de situations classiques de sujets de bac, d\u2019\u00c9duscol et de Labolyc\u00e9e : vinaigre, concentration, masse volumique, quantit\u00e9 de mati\u00e8re, dilution, titre massique.<\/p>\n<div class=\"exo\"><strong>Exercice A \u2014 vinaigre commercial.<\/strong><br>Un vinaigre commercial contient de l\u2019acide \u00e9thano\u00efque CH<sub>3<\/sub>COOH. Son titre est 8,0\u00b0, c\u2019est-\u00e0-dire 8,0 g d\u2019acide \u00e9thano\u00efque pour 100 g de vinaigre. On donne \u03c1(vinaigre)=1,01\u00d710<sup>3<\/sup> g\u00b7L<sup>\u22121<\/sup>. Calculer la concentration en quantit\u00e9 de mati\u00e8re.<div class=\"correction\"><span class=\"eq\">m<sub>acide<\/sub> = (t\/100) \u00d7 m<sub>solution<\/sub><\/span>Pour 100 g de vinaigre : m(acide)=8,0 g.<br><span class=\"eq\">n = m \/ M<\/span>n = 8,0\/60 = 0,133 mol.<br><span class=\"eq\">V = m \/ \u03c1<\/span>V = 100\/1010 = 9,90\u00d710<sup>\u22122<\/sup> L.<br><span class=\"eq\">C = n \/ V<\/span>C = 0,133\/0,0990 = <span class=\"res\">1,34 mol\u00b7L<sup>\u22121<\/sup><\/span>.<\/div><\/div>\n<div class=\"exo\"><strong>Exercice B \u2014 dilution d\u2019un vinaigre.<\/strong><br>On dilue 100 fois un vinaigre commercial. La solution dilu\u00e9e obtenue a pour concentration C<sub>d<\/sub>=2,4\u00d710<sup>\u22122<\/sup> mol\u00b7L<sup>\u22121<\/sup>. Quelle est la concentration C du vinaigre commercial ?<div class=\"correction\"><span class=\"eq\">C = facteur de dilution \u00d7 C<sub>d<\/sub><\/span>C = 100 \u00d7 2,4\u00d710<sup>\u22122<\/sup> = <span class=\"res\">2,4 mol\u00b7L<sup>\u22121<\/sup><\/span>.<\/div><\/div>\n<div class=\"exo\"><strong>Exercice C \u2014 solution d\u00e9sinfectante.<\/strong><br>Une solution contient du peroxyde d\u2019hydrog\u00e8ne H<sub>2<\/sub>O<sub>2<\/sub> \u00e0 la concentration en masse C<sub>m<\/sub>=34 g\u00b7L<sup>\u22121<\/sup>. Calculer la concentration en quantit\u00e9 de mati\u00e8re C.<div class=\"correction\">M(H<sub>2<\/sub>O<sub>2<\/sub>)=2\u00d71+2\u00d716=34 g\u00b7mol<sup>\u22121<\/sup>.<br><span class=\"eq\">C = C<sub>m<\/sub> \/ M<\/span>C = 34\/34 = <span class=\"res\">1,0 mol\u00b7L<sup>\u22121<\/sup><\/span>.<\/div><\/div>\n<div class=\"exo\"><strong>Exercice D \u2014 pr\u00e9paration par dissolution.<\/strong><br>On veut pr\u00e9parer V = 250,0 mL d\u2019une solution de chlorure de sodium de concentration C = 0,150 mol\u00b7L<sup>\u22121<\/sup>. Quelle masse de NaCl faut-il peser ?<div class=\"correction\">V = 250,0 mL = 250,0\u00d710<sup>\u22123<\/sup> L = 0,2500 L.<br><span class=\"eq\">n = C \u00d7 V<\/span>n = 0,150\u00d70,2500 = 3,75\u00d710<sup>\u22122<\/sup> mol.<br>M(NaCl)=58,5 g\u00b7mol<sup>\u22121<\/sup>.<br><span class=\"eq\">m = n \u00d7 M<\/span>m = 3,75\u00d710<sup>\u22122<\/sup>\u00d758,5 = <span class=\"res\">2,19 g<\/span>.<\/div><\/div>\n<\/section>\n\n<section id=\"carte\" class=\"card\">\n<h2>9. Carte mentale \u2014 choisir la bonne formule<\/h2>\n<div class=\"mental\">\n<div class=\"line l1\"><\/div><div class=\"line l2\"><\/div><div class=\"line l3\"><\/div><div class=\"line l4\"><\/div><div class=\"line l5\"><\/div>\n<div class=\"node center\"><span class=\"title\">Quantit\u00e9 de mati\u00e8re<\/span><strong>n<\/strong><br>unit\u00e9 : mol<\/div>\n<div class=\"node solid\"><span class=\"title\">Solide pur<\/span>Donn\u00e9e : masse m<br><span class=\"red\">n = m \/ M<\/span><br><small>m en g, M en g\u00b7mol<sup>\u22121<\/sup><\/small><\/div>\n<div class=\"node liquid\"><span class=\"title\">Liquide pur<\/span>Donn\u00e9e : volume V<br><span class=\"red\">m = \u03c1V<\/span><br>puis <span class=\"red\">n = m\/M<\/span><\/div>\n<div class=\"node solution\"><span class=\"title\">Solution<\/span>Donn\u00e9e : concentration<br><span class=\"red\">n = C \u00d7 V<\/span><br>ou <span class=\"red\">m = C<sub>m<\/sub>V<\/span><\/div>\n<div class=\"node gas\"><span class=\"title\">Gaz<\/span>Donn\u00e9e : volume du gaz<br><span class=\"red\">n = V \/ V<sub>m<\/sub><\/span><br><small>V en L<\/small><\/div>\n<div class=\"node teneur\"><span class=\"title\">Teneur massique<\/span><span class=\"red\">t(%) = m<sub>solut\u00e9<\/sub>\/m<sub>solution<\/sub> \u00d7100<\/span><br><strong>Astuce :<\/strong> raisonner sur 1,00 L de solution, puis utiliser \u03c1.<\/div>\n<\/div>\n<\/section>\n\n<section id=\"resume\" class=\"card\">\n<h2>10. R\u00e9sum\u00e9 final<\/h2>\n<div class=\"two\">\n<p><span class=\"formula\">N = nN<sub>A<\/sub><\/span><\/p>\n<p><span class=\"formula\">n = m\/M<\/span><\/p>\n<p><span class=\"formula\">m = \u03c1V<\/span><\/p>\n<p><span class=\"formula\">\u03c1 = d\u00d71000<\/span><\/p>\n<p><span class=\"formula\">n = V\/V<sub>m<\/sub><\/span><\/p>\n<p><span class=\"formula\">C = n\/V<\/span><\/p>\n<p><span class=\"formula\">C<sub>m<\/sub> = m\/V<\/span><\/p>\n<p><span class=\"formula\">C = C<sub>m<\/sub>\/M<\/span><\/p>\n<p><span class=\"formula\">t(%) = m<sub>solut\u00e9<\/sub>\/m<sub>solution<\/sub> \u00d7100<\/span><\/p>\n<\/div>\n<div class=\"warn\"><strong>Pi\u00e8ge principal :<\/strong> presque toutes les erreurs viennent des unit\u00e9s. Avant de calculer, convertir les volumes en litres : par exemple 50 mL = 50\u00d710<sup>\u22123<\/sup> L = 5,0\u00d710<sup>\u22122<\/sup> L.<\/div>\n<h3>Quiz rapide<\/h3>\n<ol><li>Quelle est l\u2019unit\u00e9 de n ?<\/li><li>Quelle formule utiliser pour un solide pur de masse connue ?<\/li><li>Que signifie un vinaigre \u00e0 6,0 % en masse ?<\/li><li>Pourquoi \u00e9crit-on 50 mL = 50\u00d710<sup>\u22123<\/sup> L ?<\/li><li>Quelle est la diff\u00e9rence entre C et C<sub>m<\/sub> ?<\/li><\/ol>\n<\/section>\n<\/div>\n\n<section class=\"audioText\">\n<h2>Texte lu par le bouton \u00ab Lire le cours \u00bb<\/h2>\n<p>Le bouton en haut de page lit automatiquement le contenu principal du cours avec la voix disponible dans le navigateur. Pour une meilleure \u00e9coute, utilise Safari ou Chrome \u00e0 jour.<\/p>\n<\/section>\n\n<p class=\"footer\">CH1 Chimie \u2014 Quantit\u00e9 de mati\u00e8re et solutions \u00b7 Premi\u00e8re sp\u00e9cialit\u00e9 physique-chimie<\/p>\n<\/main>\n<script>\nlet utterance = null;\nfunction cleanText(el){\n return el.innerText.replace(\/\\s+\/g,' ').replace(\/\u00d7\/g,' fois ').replace(\/\u03c1\/g,' rho ').replace(\/\u00b7\/g,' ').trim();\n}\nfunction speakCourse(){\n if(!('speechSynthesis' in window)){ alert('La lecture vocale n\u2019est pas disponible sur ce navigateur.'); return; }\n window.speechSynthesis.cancel();\n const text = cleanText(document.getElementById('courseText'));\n utterance = new SpeechSynthesisUtterance(text);\n utterance.lang = 'fr-FR';\n utterance.rate = 0.92;\n utterance.pitch = 1;\n const voices = window.speechSynthesis.getVoices();\n const frVoice = voices.find(v => v.lang && v.lang.toLowerCase().startsWith('fr'));\n if(frVoice) utterance.voice = frVoice;\n window.speechSynthesis.speak(utterance);\n}\nfunction pauseSpeech(){\n if(!('speechSynthesis' in window)) return;\n if(window.speechSynthesis.paused){ window.speechSynthesis.resume(); }\n else { window.speechSynthesis.pause(); }\n}\nfunction stopSpeech(){ if('speechSynthesis' in window) window.speechSynthesis.cancel(); }\nwindow.speechSynthesis && window.speechSynthesis.onvoiceschanged;\n<\/script>\n\n<h2 id=\"audio-summary\" style=\"display:none;\">R\u00e9sum\u00e9 audio du chapitre<\/h2>\n<div id=\"audioText\" style=\"display:none;\">\nBienvenue dans le chapitre quantit\u00e9 de mati\u00e8re et solutions.\n\nLa quantit\u00e9 de mati\u00e8re se note n et s\u2019exprime en mole.\nUne mole contient 6 virgule 02 fois 10 puissance 23 entit\u00e9s chimiques.\nLa relation fondamentale est :\nN \u00e9gal n fois N A.\n\nPour un solide pur :\nn \u00e9gal m sur M.\nPetit m est la masse en grammes.\nGrand M est la masse molaire en grammes par mole.\n\nPour un liquide pur :\non utilise d\u2019abord la masse volumique.\nRho \u00e9gal m sur V.\nPuis :\nn \u00e9gal m sur M.\n\nPour un gaz :\nn \u00e9gal V sur V m.\nAvec V m le volume molaire.\n\u00c0 20 degr\u00e9s Celsius et pression atmosph\u00e9rique,\nV m vaut environ 24 litres par mole.\n\nPour une solution :\nla concentration molaire se note C.\nLa relation est :\nC \u00e9gal n sur V.\n\nAttention aux conversions :\nun millilitre correspond \u00e0 10 puissance moins 3 litre.\n\nPour la teneur massique :\nune solution \u00e0 6 pourcent signifie 6 grammes de solut\u00e9 pour 100 grammes de solution.\n\nAstuce importante :\nraisonner sur un litre de solution permet souvent de simplifier les calculs.\nOn utilise alors la masse volumique pour obtenir la masse totale de solution.\n\nM\u00e9thode type :\nidentifier les donn\u00e9es,\n\u00e9crire les formules,\neffectuer les conversions,\nremplacer les valeurs,\n\u00e9crire le r\u00e9sultat avec les unit\u00e9s.\n\nFin du r\u00e9sum\u00e9 du chapitre.\n<\/div>\n\n<div style=\"margin:20px 0;text-align:center;\">\n<button onclick=\"playAudioSummary()\" style=\"background:#c62828;color:white;border:none;padding:14px 24px;border-radius:10px;font-size:18px;cursor:pointer;\">\n\ud83d\udd0a \u00c9couter le r\u00e9sum\u00e9 audio du chapitre\n<\/button>\n<button onclick=\"stopAudioSummary()\" style=\"background:#333;color:white;border:none;padding:14px 24px;border-radius:10px;font-size:18px;cursor:pointer;margin-left:10px;\">\n\u23f9 Arr\u00eater\n<\/button>\n<\/div>\n\n<script>\nlet speechSynthesisUtterance;\n\nfunction playAudioSummary() {\n const text = document.getElementById(\"audioText\").innerText;\n\n speechSynthesis.cancel();\n\n speechSynthesisUtterance = new SpeechSynthesisUtterance(text);\n speechSynthesisUtterance.lang = 'fr-FR';\n speechSynthesisUtterance.rate = 0.92;\n speechSynthesisUtterance.pitch = 1;\n\n const voices = speechSynthesis.getVoices();\n const frenchVoice = voices.find(v => v.lang.includes('fr'));\n if (frenchVoice) {\n speechSynthesisUtterance.voice = frenchVoice;\n }\n\n speechSynthesis.speak(speechSynthesisUtterance);\n}\n\nfunction stopAudioSummary() {\n speechSynthesis.cancel();\n}\n<\/script>\n\n<\/body>\n<\/html>\n\n","protected":false},"excerpt":{"rendered":"<p>CH1 Chimie \u2014 Quantit\u00e9 de mati\u00e8re et solutions Premi\u00e8re sp\u00e9cialit\u00e9 physique-chimie \u00b7 Chimie CH1 \u2014 Quantit\u00e9 de mati\u00e8re et solutions Objectif : passer des grandeurs mesurables...<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-711","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages\/711","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/comments?post=711"}],"version-history":[{"count":2,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages\/711\/revisions"}],"predecessor-version":[{"id":715,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages\/711\/revisions\/715"}],"wp:attachment":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/media?parent=711"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}

{"id":711,"date":"2026-05-26T12:21:11","date_gmt":"2026-05-26T10:21:11","guid":{"rendered":"https:\/\/pcwallis.malo.wf\/?page_id=711"},"modified":"2026-05-26T12:29:34","modified_gmt":"2026-05-26T10:29:34","slug":"ch1-quantite-de-matiere-et-solutions","status":"publish","type":"page","link":"https:\/\/pcwallis.malo.wf\/index.php\/ch1-quantite-de-matiere-et-solutions\/","title":{"rendered":"ch1 quantit\u00e9 de mati\u00e8re et solutions"},"content":{"rendered":"\n\n\n\n\n\nCH1 Chimie \u2014 Quantit\u00e9 de mati\u00e8re et solutions<\/title>\n<style>\n:root{--blue:#113d72;--blue2:#1763aa;--cyan:#eaf6ff;--red:#c62828;--redbg:#fff2f2;--green:#14734d;--greenbg:#edfdf5;--ink:#142033;--muted:#5f6f85;--paper:#fff;--bg:#f4f8ff;--line:#d8e2ef;--yellow:#fff7d7;--violet:#6d28d9;--orange:#f97316}\n*{box-sizing:border-box}html{scroll-behavior:smooth}body{margin:0;background:linear-gradient(135deg,#f6f9ff,#eef6ff);color:var(--ink);font-family:Inter,Segoe 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#ccd5e1;margin-bottom:10px}.red{color:#c62828!important}.formula,.eq{border-color:#c62828!important;color:#c62828!important}.mental{break-inside:avoid}}\n<\/style>\n<\/head>\n<body>\n<main class=\"page\">\n<section class=\"hero\">\n<span class=\"tag\">Premi\u00e8re sp\u00e9cialit\u00e9 physique-chimie \u00b7 Chimie<\/span>\n<h1>CH1 \u2014 Quantit\u00e9 de mati\u00e8re et solutions<\/h1>\n<p>Objectif : passer des grandeurs mesurables au laboratoire \u2014 masse, volume, masse volumique, concentration \u2014 \u00e0 la quantit\u00e9 de mati\u00e8re <strong>n<\/strong>, exprim\u00e9e en mole.<\/p>\n<p><span class=\"pill\">mole<\/span><span class=\"pill\">masse molaire<\/span><span class=\"pill\">volume molaire<\/span><span class=\"pill\">solution<\/span><span class=\"pill\">concentration<\/span><span class=\"pill\">teneur massique<\/span><\/p>\n<div class=\"toolbar\">\n<button class=\"btn\" onclick=\"speakCourse()\">\u25b6 Lire le cours<\/button>\n<button class=\"btn secondary\" onclick=\"pauseSpeech()\">\u23f8 Pause \/ reprise<\/button>\n<button class=\"btn stop\" onclick=\"stopSpeech()\">\u25a0 Stop<\/button>\n<\/div>\n<div class=\"voiceNote\">Lecture vocale par le navigateur : fonctionne avec Safari, Chrome, Edge, Firefox selon les voix disponibles.<\/div>\n<\/section>\n\n<section class=\"grid\">\n<div class=\"card span4 toc\">\n<h2>Plan<\/h2>\n<a href=\"#mole\">1. Mole et constante d\u2019Avogadro<\/a>\n<a href=\"#masse-molaire\">2. Masse molaire<\/a>\n<a href=\"#calculs\">3. Solide pur<\/a>\n<a href=\"#liquides\">4. Liquide pur<\/a>\n<a href=\"#solutions\">5. Solutions<\/a>\n<a href=\"#teneur\">6. Teneur massique<\/a>\n<a href=\"#gaz\">7. Gaz<\/a>\n<a href=\"#exercices\">8. Exercices type bac<\/a>\n<a href=\"#carte\">9. Carte mentale<\/a>\n<a href=\"#resume\">10. R\u00e9sum\u00e9 + quiz<\/a>\n<\/div>\n<div class=\"card span8\">\n<h2>Capacit\u00e9s travaill\u00e9es<\/h2>\n<ul>\n<li>D\u00e9terminer une masse molaire \u00e0 partir des masses molaires atomiques.<\/li>\n<li>D\u00e9terminer une quantit\u00e9 de mati\u00e8re \u00e0 partir d\u2019une masse.<\/li>\n<li>Utiliser une masse volumique ou une densit\u00e9 pour passer d\u2019un volume \u00e0 une masse.<\/li>\n<li>Utiliser le volume molaire d\u2019un gaz.<\/li>\n<li>D\u00e9terminer une quantit\u00e9 de mati\u00e8re \u00e0 partir d\u2019une concentration et d\u2019un volume de solution.<\/li>\n<\/ul>\n<div class=\"warn\"><strong>Fil rouge du chapitre :<\/strong> identifier la situation, choisir la bonne formule rouge, convertir les unit\u00e9s, puis souligner le r\u00e9sultat avec son unit\u00e9.<\/div>\n<\/div>\n<\/section>\n\n<div id=\"courseText\">\n<section id=\"mole\" class=\"card\">\n<h2>1. La mole : compter les entit\u00e9s chimiques<\/h2>\n<p>Pour compter les mol\u00e9cules, les atomes ou les ions, on utilise un paquet de r\u00e9f\u00e9rence : <strong>la mole<\/strong>.<\/p>\n<div class=\"greenbox\">\n<p><strong>Rappel :<\/strong> <span class=\"red\">1 mole contient<\/span> <strong>N<sub>A<\/sub> = 6,02 \u00d7 10<sup>23<\/sup> entit\u00e9s\u00b7mol<sup>\u22121<\/sup><\/strong>.<\/p>\n<p>N<sub>A<\/sub> est la <strong>constante d\u2019Avogadro<\/strong>.<\/p>\n<\/div>\n<p><span class=\"formula\">N = n \u00d7 N<sub>A<\/sub><\/span> <span class=\"formula blue\">n = N \/ N<sub>A<\/sub><\/span><\/p>\n<table><tr><th>Grandeur<\/th><th>Signification<\/th><th>Unit\u00e9<\/th><\/tr><tr><td>N<\/td><td>nombre d\u2019entit\u00e9s : atomes, ions, mol\u00e9cules<\/td><td>sans unit\u00e9<\/td><\/tr><tr><td>n<\/td><td>quantit\u00e9 de mati\u00e8re<\/td><td>mol<\/td><\/tr><tr><td>N<sub>A<\/sub><\/td><td>constante d\u2019Avogadro<\/td><td>mol<sup>\u22121<\/sup><\/td><\/tr><\/table>\n<div class=\"exo\"><strong>Exemple corrig\u00e9.<\/strong> Si on pr\u00e9l\u00e8ve n = 1,0 \u00d7 10<sup>2<\/sup> mol de cuivre :<div class=\"correction\">N = n \u00d7 N<sub>A<\/sub> = 1,0 \u00d7 10<sup>2<\/sup> \u00d7 6,02 \u00d7 10<sup>23<\/sup> = <span class=\"res\">6,02 \u00d7 10<sup>25<\/sup> atomes<\/span>.<\/div><\/div>\n<\/section>\n\n<section id=\"masse-molaire\" class=\"card\">\n<h2>2. Masse molaire<\/h2>\n<h3>2.1 Masse molaire atomique<\/h3>\n<p>La <strong>masse molaire atomique<\/strong> d\u2019un \u00e9l\u00e9ment est la masse d\u2019une mole d\u2019atomes de cet \u00e9l\u00e9ment.<\/p>\n<div class=\"redbox\"><p><span class=\"red\">\u00c0 retenir :<\/span> une masse molaire s\u2019exprime en <strong>g\u00b7mol<sup>\u22121<\/sup><\/strong>.<\/p><\/div>\n<p>Exemple : M(O) = 16 g\u00b7mol<sup>\u22121<\/sup>.<\/p>\n<h3>2.2 Masse molaire mol\u00e9culaire<\/h3>\n<p>Pour une mol\u00e9cule, on additionne les masses molaires atomiques de tous les atomes pr\u00e9sents.<\/p>\n<p><span class=\"formula\">M(C<sub>x<\/sub>H<sub>y<\/sub>O<sub>z<\/sub>) = xM(C) + yM(H) + zM(O)<\/span><\/p>\n<div class=\"bluebox\"><strong>Exemple : glucose C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub><\/strong><br>M = 6\u00d712 + 12\u00d71 + 6\u00d716 = <span class=\"res\">180 g\u00b7mol<sup>\u22121<\/sup><\/span>.<\/div>\n<div class=\"bluebox\"><strong>Exemple : eau H<sub>2<\/sub>O<\/strong><br>M = 2\u00d71 + 16 = <span class=\"res\">18 g\u00b7mol<sup>\u22121<\/sup><\/span>.<\/div>\n<h3>2.3 Masse molaire ionique<\/h3>\n<p>Pour un ion, <span class=\"red\">on ne tient pas compte de la charge \u00e9lectrique<\/span> dans le calcul de la masse molaire.<\/p>\n<p>Exemple : M(ClO<sup>\u2212<\/sup>) = M(Cl) + M(O).<\/p>\n<\/section>\n\n<section id=\"calculs\" class=\"card\">\n<h2>3. Calculer une quantit\u00e9 de mati\u00e8re \u00e0 partir d\u2019un solide pur<\/h2>\n<p><span class=\"formula\">n = m \/ M<\/span> <span class=\"formula blue\">m = n \u00d7 M<\/span><\/p>\n<table><tr><th>Grandeur<\/th><th>Unit\u00e9 \u00e0 utiliser<\/th><\/tr><tr><td>n<\/td><td>mol<\/td><\/tr><tr><td>m<\/td><td>g<\/td><\/tr><tr><td>M<\/td><td>g\u00b7mol<sup>\u22121<\/sup><\/td><\/tr><\/table>\n<div class=\"exo\"><strong>Exercice type 1 \u2014 glucose.<\/strong> On pr\u00e9l\u00e8ve m = 2,0 g de glucose C<sub>6<\/sub>H<sub>12<\/sub>O<sub>6<\/sub>. Calculer n.<div class=\"correction\"><span class=\"eq\">n = m \/ M<\/span>M = 180 g\u00b7mol<sup>\u22121<\/sup><br>n = 2,0 \/ 180 = <span class=\"res\">1,1 \u00d7 10<sup>\u22122<\/sup> mol<\/span>.<\/div><\/div>\n<div class=\"exo\"><strong>Exercice type 2 \u2014 chlorure de sodium.<\/strong> Calculer la masse \u00e0 peser pour pr\u00e9lever n = 2,0\u00d710<sup>\u22121<\/sup> mol de NaCl(s). Donn\u00e9es : M(Na)=23,0 g\u00b7mol<sup>\u22121<\/sup>, M(Cl)=35,5 g\u00b7mol<sup>\u22121<\/sup>.<div class=\"correction\">M(NaCl)=23,0+35,5=58,5 g\u00b7mol<sup>\u22121<\/sup>.<br><span class=\"eq\">m = n \u00d7 M<\/span>m = 2,0\u00d710<sup>\u22121<\/sup> \u00d7 58,5 = <span class=\"res\">12 g<\/span>.<\/div><\/div>\n<\/section>\n\n<section id=\"liquides\" class=\"card\">\n<h2>4. Calculer une quantit\u00e9 de mati\u00e8re \u00e0 partir du volume d\u2019un liquide pur<\/h2>\n<p>Pour un liquide pur, on ne peut pas utiliser directement le volume dans <span class=\"red\">n = m \/ M<\/span>. Il faut d\u2019abord trouver la masse du liquide.<\/p>\n<h3>4.1 Avec la masse volumique \u03c1<\/h3>\n<p><span class=\"formula\">\u03c1 = m \/ V<\/span> donc <span class=\"formula blue\">m = \u03c1 \u00d7 V<\/span><\/p>\n<div class=\"warn\"><strong>Unit\u00e9s coh\u00e9rentes :<\/strong> si \u03c1 est en g\u00b7L<sup>\u22121<\/sup>, alors V doit \u00eatre en L et m sera en g.<\/div>\n<h3>4.2 Avec la densit\u00e9 d<\/h3>\n<p>Pour un liquide, la densit\u00e9 compare sa masse volumique \u00e0 celle de l\u2019eau.<\/p>\n<p><span class=\"formula\">d = \u03c1<sub>liquide<\/sub> \/ \u03c1<sub>eau<\/sub><\/span> avec <strong>\u03c1<sub>eau<\/sub> = 1000 g\u00b7L<sup>\u22121<\/sup><\/strong>.<\/p>\n<p>Donc : <span class=\"formula\">\u03c1<sub>liquide<\/sub> = d \u00d7 1000<\/span><\/p>\n<div class=\"exo\"><strong>Exercice type 3 \u2014 \u00e9thanol.<\/strong> D\u00e9terminer la quantit\u00e9 de mati\u00e8re dans V = 10 mL d\u2019\u00e9thanol pur. Donn\u00e9es : \u03c1 = 803 g\u00b7L<sup>\u22121<\/sup> ; M(\u00e9thanol)=46 g\u00b7mol<sup>\u22121<\/sup>.<div class=\"correction\"><span class=\"red\">Conversion :<\/span> V = 10 mL = 10 \u00d7 10<sup>\u22123<\/sup> L = 1,0\u00d710<sup>\u22122<\/sup> L.<br><span class=\"eq\">m = \u03c1 \u00d7 V<\/span>m = 803 \u00d7 1,0\u00d710<sup>\u22122<\/sup> = 8,03 g.<br><span class=\"eq\">n = m \/ M<\/span>n = 8,03 \/ 46 = <span class=\"res\">1,7\u00d710<sup>\u22121<\/sup> mol<\/span>.<\/div><\/div>\n<div class=\"exo\"><strong>Exercice type 4 \u2014 acide sulfurique.<\/strong> Calculer la quantit\u00e9 de mati\u00e8re dans V = 20 mL d\u2019acide sulfurique H<sub>2<\/sub>SO<sub>4<\/sub> pur de densit\u00e9 d = 1,8. Donn\u00e9e : M(H<sub>2<\/sub>SO<sub>4<\/sub>) = 98 g\u00b7mol<sup>\u22121<\/sup>.<div class=\"correction\"><span class=\"eq\">\u03c1 = d \u00d7 1000<\/span>\u03c1 = 1,8 \u00d7 1000 = 1800 g\u00b7L<sup>\u22121<\/sup>.<br>V = 20 mL = 20\u00d710<sup>\u22123<\/sup> L = 2,0\u00d710<sup>\u22122<\/sup> L.<br><span class=\"eq\">m = \u03c1 \u00d7 V<\/span>m = 1800 \u00d7 2,0\u00d710<sup>\u22122<\/sup> = 36 g.<br><span class=\"eq\">n = m \/ M<\/span>n = 36 \/ 98 = <span class=\"res\">3,7\u00d710<sup>\u22121<\/sup> mol<\/span>.<\/div><\/div>\n<\/section>\n\n<section id=\"solutions\" class=\"card\">\n<h2>5. Solutions : concentration en masse et concentration en quantit\u00e9 de mati\u00e8re<\/h2>\n<p>Une solution contient un <strong>solvant<\/strong> et un ou plusieurs <strong>solut\u00e9s<\/strong>. On utilise deux concentrations importantes.<\/p>\n<table><tr><th>Nom<\/th><th>Formule rouge<\/th><th>Unit\u00e9s<\/th><th>Signification<\/th><\/tr><tr><td>Concentration en masse<\/td><td><span class=\"red\">C<sub>m<\/sub> = m \/ V<\/span><\/td><td>g\u00b7L<sup>\u22121<\/sup><\/td><td>masse de solut\u00e9 par litre de solution<\/td><\/tr><tr><td>Concentration en quantit\u00e9 de mati\u00e8re<\/td><td><span class=\"red\">C = n \/ V<\/span><\/td><td>mol\u00b7L<sup>\u22121<\/sup><\/td><td>quantit\u00e9 de mati\u00e8re de solut\u00e9 par litre de solution<\/td><\/tr><\/table>\n<p><span class=\"formula\">n = C \u00d7 V<\/span> <span class=\"formula blue\">m = C<sub>m<\/sub> \u00d7 V<\/span> <span class=\"formula green\">C = C<sub>m<\/sub> \/ M<\/span><\/p>\n<div class=\"warn\"><strong>Volume en litres obligatoire :<\/strong><br>50 mL = <span class=\"red\">50 \u00d7 10<sup>\u22123<\/sup> L<\/span> = <span class=\"red\">5,0 \u00d7 10<sup>\u22122<\/sup> L<\/span>.<br>On \u00e9crit d\u2019abord \u00ab \u00d710<sup>\u22123<\/sup> \u00bb pour bien retenir que <strong>1 mL = 10<sup>\u22123<\/sup> L<\/strong>.<\/div>\n<div class=\"exo\"><strong>Exercice type 5 \u2014 solution de glucose.<\/strong> Une solution contient 9,0 g de glucose dans V = 250 mL. Calculer C<sub>m<\/sub>, puis C.<div class=\"correction\"><span class=\"red\">Conversion :<\/span> V = 250 mL = 250\u00d710<sup>\u22123<\/sup> L = 0,250 L.<br><span class=\"eq\">C<sub>m<\/sub> = m \/ V<\/span>C<sub>m<\/sub> = 9,0 \/ 0,250 = <span class=\"res\">36 g\u00b7L<sup>\u22121<\/sup><\/span>.<br>M(glucose)=180 g\u00b7mol<sup>\u22121<\/sup>.<br><span class=\"eq\">C = C<sub>m<\/sub> \/ M<\/span>C = 36 \/ 180 = <span class=\"res\">0,20 mol\u00b7L<sup>\u22121<\/sup><\/span>.<\/div><\/div>\n<\/section>\n\n<section id=\"teneur\" class=\"card\">\n<h2>6. Teneur massique, pourcentage massique et vinaigre<\/h2>\n<p>La <strong>teneur massique<\/strong>, ou <strong>pourcentage massique<\/strong>, indique la masse de solut\u00e9 contenue dans une masse donn\u00e9e de solution.<\/p>\n<div class=\"redbox\"><p><span class=\"red\">Formule :<\/span><\/p><p><span class=\"formula\">t = m<sub>solut\u00e9<\/sub> \/ m<sub>solution<\/sub><\/span><\/p><p>Si on l\u2019exprime en pourcentage :<\/p><p><span class=\"formula\">t(%) = (m<sub>solut\u00e9<\/sub> \/ m<sub>solution<\/sub>) \u00d7 100<\/span><\/p><\/div>\n<div class=\"greenbox\"><strong>Interpr\u00e9tation :<\/strong> un vinaigre \u00e0 6,0 % en masse signifie <span class=\"red\">6,0 g d\u2019acide \u00e9thano\u00efque pour 100 g de vinaigre<\/span>.<\/div>\n<div class=\"warn\"><strong>Astuce tr\u00e8s efficace :<\/strong> raisonner sur <strong>1,00 L de solution<\/strong>. On calcule alors la masse d\u2019un litre de vinaigre avec la masse volumique, puis on applique le pourcentage massique.<\/div>\n<p>Donn\u00e9e utilis\u00e9e ici : <strong>\u03c1(vinaigre) = 1,01\u00d710<sup>3<\/sup> g\u00b7L<sup>\u22121<\/sup><\/strong>. Cette valeur est r\u00e9aliste pour un vinaigre alimentaire et permet de faire un calcul plus complet.<\/p>\n\n<div class=\"exo\"><strong>Exercice type \u2014 concentration d\u2019un vinaigre \u00e0 6,0 %.<\/strong><br>Un vinaigre contient 6,0 % en masse d\u2019acide \u00e9thano\u00efque CH<sub>3<\/sub>COOH. Sa masse volumique vaut \u03c1 = 1,01\u00d710<sup>3<\/sup> g\u00b7L<sup>\u22121<\/sup>. Calculer la concentration en quantit\u00e9 de mati\u00e8re C en acide \u00e9thano\u00efque.<div class=\"correction\">\n<h4>M\u00e9thode 1 \u2014 avec calculs interm\u00e9diaires<\/h4>\n<p><span class=\"red\">On raisonne sur V = 1,00 L de vinaigre.<\/span><\/p>\n<span class=\"eq\">m<sub>solution<\/sub> = \u03c1 \u00d7 V<\/span>\n<p>m<sub>solution<\/sub> = 1,01\u00d710<sup>3<\/sup> \u00d7 1,00 = 1,01\u00d710<sup>3<\/sup> g = 1010 g.<\/p>\n<span class=\"eq\">m<sub>solut\u00e9<\/sub> = (t(%) \/ 100) \u00d7 m<sub>solution<\/sub><\/span>\n<p>m<sub>acide<\/sub> = (6,0 \/ 100) \u00d7 1010 = 60,6 g.<\/p>\n<p>M(CH<sub>3<\/sub>COOH) = 2\u00d712 + 4\u00d71 + 2\u00d716 = 60 g\u00b7mol<sup>\u22121<\/sup>.<\/p>\n<span class=\"eq\">n = m \/ M<\/span>\n<p>n = 60,6 \/ 60 = 1,01 mol.<\/p>\n<span class=\"eq\">C = n \/ V<\/span>\n<p>C = 1,01 \/ 1,00 = <span class=\"res\">1,01 mol\u00b7L<sup>\u22121<\/sup><\/span>.<\/p>\n<h4>M\u00e9thode 2 \u2014 r\u00e9solution litt\u00e9rale<\/h4>\n<p><strong>\u00c9quation 1 :<\/strong> <span class=\"red\">m<sub>solution<\/sub> = \u03c1V<\/span><\/p>\n<p><strong>\u00c9quation 2 :<\/strong> <span class=\"red\">m<sub>solut\u00e9<\/sub> = (t\/100) \u00d7 m<sub>solution<\/sub><\/span><\/p>\n<p><strong>\u00c9quation 3 :<\/strong> <span class=\"red\">n = m<sub>solut\u00e9<\/sub> \/ M<\/span><\/p>\n<p><strong>\u00c9quation 4 :<\/strong> <span class=\"red\">C = n \/ V<\/span><\/p>\n<p>En injectant 1 dans 2, puis 2 dans 3, puis 3 dans 4 :<\/p>\n<span class=\"eq\">C = [(t\/100) \u00d7 \u03c1 \u00d7 V] \/ (M \u00d7 V)<\/span>\n<p>Le volume V se simplifie :<\/p>\n<span class=\"eq\">C = (t\/100) \u00d7 \u03c1 \/ M<\/span>\n<p>C = (6,0\/100) \u00d7 1,01\u00d710<sup>3<\/sup> \/ 60 = <span class=\"res\">1,01 mol\u00b7L<sup>\u22121<\/sup><\/span>.<\/p>\n<\/div><\/div>\n<\/section>\n\n<section id=\"gaz\" class=\"card\">\n<h2>7. Quantit\u00e9 de mati\u00e8re d\u2019un gaz : volume molaire<\/h2>\n<p>Le <strong>volume molaire<\/strong> V<sub>m<\/sub> est le volume occup\u00e9 par une mole de gaz dans des conditions donn\u00e9es de temp\u00e9rature et de pression.<\/p>\n<div class=\"redbox\"><p><span class=\"red\">\u00c0 25 \u00b0C et sous pression atmosph\u00e9rique :<\/span> V<sub>m<\/sub> \u2248 <strong>22,4 L\u00b7mol<sup>\u22121<\/sup><\/strong> dans ce cours.<\/p><\/div>\n<p><span class=\"formula\">n = V \/ V<sub>m<\/sub><\/span> <span class=\"formula blue\">V = n \u00d7 V<sub>m<\/sub><\/span><\/p>\n<div class=\"exo\"><strong>Exercice type 6 \u2014 dioxyg\u00e8ne.<\/strong> Une r\u00e9action produit n = 1,0\u00d710<sup>\u22122<\/sup> mol de dioxyg\u00e8ne O<sub>2<\/sub>. Quel volume de gaz est d\u00e9gag\u00e9 ?<div class=\"correction\"><span class=\"eq\">V = n \u00d7 V<sub>m<\/sub><\/span>V = 1,0\u00d710<sup>\u22122<\/sup> \u00d7 22,4 = <span class=\"res\">2,24\u00d710<sup>\u22121<\/sup> L<\/span>, soit <span class=\"res\">224 mL<\/span>.<\/div><\/div>\n<div class=\"exo\"><strong>Exercice type 7 \u2014 r\u00e9servoir.<\/strong> Un r\u00e9servoir contient V = 50 L de gaz \u00e0 pression atmosph\u00e9rique. Calculer la quantit\u00e9 de mati\u00e8re.<div class=\"correction\"><span class=\"eq\">n = V \/ V<sub>m<\/sub><\/span>n = 50 \/ 22,4 = <span class=\"res\">2,2 mol<\/span>.<\/div><\/div>\n<\/section>\n\n<section id=\"exercices\" class=\"card\">\n<h2>8. Exercices type bac \u2014 entra\u00eenement<\/h2>\n<p class=\"small\">Exercices inspir\u00e9s de situations classiques de sujets de bac, d\u2019\u00c9duscol et de Labolyc\u00e9e : vinaigre, concentration, masse volumique, quantit\u00e9 de mati\u00e8re, dilution, titre massique.<\/p>\n<div class=\"exo\"><strong>Exercice A \u2014 vinaigre commercial.<\/strong><br>Un vinaigre commercial contient de l\u2019acide \u00e9thano\u00efque CH<sub>3<\/sub>COOH. Son titre est 8,0\u00b0, c\u2019est-\u00e0-dire 8,0 g d\u2019acide \u00e9thano\u00efque pour 100 g de vinaigre. On donne \u03c1(vinaigre)=1,01\u00d710<sup>3<\/sup> g\u00b7L<sup>\u22121<\/sup>. Calculer la concentration en quantit\u00e9 de mati\u00e8re.<div class=\"correction\"><span class=\"eq\">m<sub>acide<\/sub> = (t\/100) \u00d7 m<sub>solution<\/sub><\/span>Pour 100 g de vinaigre : m(acide)=8,0 g.<br><span class=\"eq\">n = m \/ M<\/span>n = 8,0\/60 = 0,133 mol.<br><span class=\"eq\">V = m \/ \u03c1<\/span>V = 100\/1010 = 9,90\u00d710<sup>\u22122<\/sup> L.<br><span class=\"eq\">C = n \/ V<\/span>C = 0,133\/0,0990 = <span class=\"res\">1,34 mol\u00b7L<sup>\u22121<\/sup><\/span>.<\/div><\/div>\n<div class=\"exo\"><strong>Exercice B \u2014 dilution d\u2019un vinaigre.<\/strong><br>On dilue 100 fois un vinaigre commercial. La solution dilu\u00e9e obtenue a pour concentration C<sub>d<\/sub>=2,4\u00d710<sup>\u22122<\/sup> mol\u00b7L<sup>\u22121<\/sup>. Quelle est la concentration C du vinaigre commercial ?<div class=\"correction\"><span class=\"eq\">C = facteur de dilution \u00d7 C<sub>d<\/sub><\/span>C = 100 \u00d7 2,4\u00d710<sup>\u22122<\/sup> = <span class=\"res\">2,4 mol\u00b7L<sup>\u22121<\/sup><\/span>.<\/div><\/div>\n<div class=\"exo\"><strong>Exercice C \u2014 solution d\u00e9sinfectante.<\/strong><br>Une solution contient du peroxyde d\u2019hydrog\u00e8ne H<sub>2<\/sub>O<sub>2<\/sub> \u00e0 la concentration en masse C<sub>m<\/sub>=34 g\u00b7L<sup>\u22121<\/sup>. Calculer la concentration en quantit\u00e9 de mati\u00e8re C.<div class=\"correction\">M(H<sub>2<\/sub>O<sub>2<\/sub>)=2\u00d71+2\u00d716=34 g\u00b7mol<sup>\u22121<\/sup>.<br><span class=\"eq\">C = C<sub>m<\/sub> \/ M<\/span>C = 34\/34 = <span class=\"res\">1,0 mol\u00b7L<sup>\u22121<\/sup><\/span>.<\/div><\/div>\n<div class=\"exo\"><strong>Exercice D \u2014 pr\u00e9paration par dissolution.<\/strong><br>On veut pr\u00e9parer V = 250,0 mL d\u2019une solution de chlorure de sodium de concentration C = 0,150 mol\u00b7L<sup>\u22121<\/sup>. Quelle masse de NaCl faut-il peser ?<div class=\"correction\">V = 250,0 mL = 250,0\u00d710<sup>\u22123<\/sup> L = 0,2500 L.<br><span class=\"eq\">n = C \u00d7 V<\/span>n = 0,150\u00d70,2500 = 3,75\u00d710<sup>\u22122<\/sup> mol.<br>M(NaCl)=58,5 g\u00b7mol<sup>\u22121<\/sup>.<br><span class=\"eq\">m = n \u00d7 M<\/span>m = 3,75\u00d710<sup>\u22122<\/sup>\u00d758,5 = <span class=\"res\">2,19 g<\/span>.<\/div><\/div>\n<\/section>\n\n<section id=\"carte\" class=\"card\">\n<h2>9. Carte mentale \u2014 choisir la bonne formule<\/h2>\n<div class=\"mental\">\n<div class=\"line l1\"><\/div><div class=\"line l2\"><\/div><div class=\"line l3\"><\/div><div class=\"line l4\"><\/div><div class=\"line l5\"><\/div>\n<div class=\"node center\"><span class=\"title\">Quantit\u00e9 de mati\u00e8re<\/span><strong>n<\/strong><br>unit\u00e9 : mol<\/div>\n<div class=\"node solid\"><span class=\"title\">Solide pur<\/span>Donn\u00e9e : masse m<br><span class=\"red\">n = m \/ M<\/span><br><small>m en g, M en g\u00b7mol<sup>\u22121<\/sup><\/small><\/div>\n<div class=\"node liquid\"><span class=\"title\">Liquide pur<\/span>Donn\u00e9e : volume V<br><span class=\"red\">m = \u03c1V<\/span><br>puis <span class=\"red\">n = m\/M<\/span><\/div>\n<div class=\"node solution\"><span class=\"title\">Solution<\/span>Donn\u00e9e : concentration<br><span class=\"red\">n = C \u00d7 V<\/span><br>ou <span class=\"red\">m = C<sub>m<\/sub>V<\/span><\/div>\n<div class=\"node gas\"><span class=\"title\">Gaz<\/span>Donn\u00e9e : volume du gaz<br><span class=\"red\">n = V \/ V<sub>m<\/sub><\/span><br><small>V en L<\/small><\/div>\n<div class=\"node teneur\"><span class=\"title\">Teneur massique<\/span><span class=\"red\">t(%) = m<sub>solut\u00e9<\/sub>\/m<sub>solution<\/sub> \u00d7100<\/span><br><strong>Astuce :<\/strong> raisonner sur 1,00 L de solution, puis utiliser \u03c1.<\/div>\n<\/div>\n<\/section>\n\n<section id=\"resume\" class=\"card\">\n<h2>10. R\u00e9sum\u00e9 final<\/h2>\n<div class=\"two\">\n<p><span class=\"formula\">N = nN<sub>A<\/sub><\/span><\/p>\n<p><span class=\"formula\">n = m\/M<\/span><\/p>\n<p><span class=\"formula\">m = \u03c1V<\/span><\/p>\n<p><span class=\"formula\">\u03c1 = d\u00d71000<\/span><\/p>\n<p><span class=\"formula\">n = V\/V<sub>m<\/sub><\/span><\/p>\n<p><span class=\"formula\">C = n\/V<\/span><\/p>\n<p><span class=\"formula\">C<sub>m<\/sub> = m\/V<\/span><\/p>\n<p><span class=\"formula\">C = C<sub>m<\/sub>\/M<\/span><\/p>\n<p><span class=\"formula\">t(%) = m<sub>solut\u00e9<\/sub>\/m<sub>solution<\/sub> \u00d7100<\/span><\/p>\n<\/div>\n<div class=\"warn\"><strong>Pi\u00e8ge principal :<\/strong> presque toutes les erreurs viennent des unit\u00e9s. Avant de calculer, convertir les volumes en litres : par exemple 50 mL = 50\u00d710<sup>\u22123<\/sup> L = 5,0\u00d710<sup>\u22122<\/sup> L.<\/div>\n<h3>Quiz rapide<\/h3>\n<ol><li>Quelle est l\u2019unit\u00e9 de n ?<\/li><li>Quelle formule utiliser pour un solide pur de masse connue ?<\/li><li>Que signifie un vinaigre \u00e0 6,0 % en masse ?<\/li><li>Pourquoi \u00e9crit-on 50 mL = 50\u00d710<sup>\u22123<\/sup> L ?<\/li><li>Quelle est la diff\u00e9rence entre C et C<sub>m<\/sub> ?<\/li><\/ol>\n<\/section>\n<\/div>\n\n<section class=\"audioText\">\n<h2>Texte lu par le bouton \u00ab Lire le cours \u00bb<\/h2>\n<p>Le bouton en haut de page lit automatiquement le contenu principal du cours avec la voix disponible dans le navigateur. Pour une meilleure \u00e9coute, utilise Safari ou Chrome \u00e0 jour.<\/p>\n<\/section>\n\n<p class=\"footer\">CH1 Chimie \u2014 Quantit\u00e9 de mati\u00e8re et solutions \u00b7 Premi\u00e8re sp\u00e9cialit\u00e9 physique-chimie<\/p>\n<\/main>\n<script>\nlet utterance = null;\nfunction cleanText(el){\n return el.innerText.replace(\/\\s+\/g,' ').replace(\/\u00d7\/g,' fois ').replace(\/\u03c1\/g,' rho ').replace(\/\u00b7\/g,' ').trim();\n}\nfunction speakCourse(){\n if(!('speechSynthesis' in window)){ alert('La lecture vocale n\u2019est pas disponible sur ce navigateur.'); return; }\n window.speechSynthesis.cancel();\n const text = cleanText(document.getElementById('courseText'));\n utterance = new SpeechSynthesisUtterance(text);\n utterance.lang = 'fr-FR';\n utterance.rate = 0.92;\n utterance.pitch = 1;\n const voices = window.speechSynthesis.getVoices();\n const frVoice = voices.find(v => v.lang && v.lang.toLowerCase().startsWith('fr'));\n if(frVoice) utterance.voice = frVoice;\n window.speechSynthesis.speak(utterance);\n}\nfunction pauseSpeech(){\n if(!('speechSynthesis' in window)) return;\n if(window.speechSynthesis.paused){ window.speechSynthesis.resume(); }\n else { window.speechSynthesis.pause(); }\n}\nfunction stopSpeech(){ if('speechSynthesis' in window) window.speechSynthesis.cancel(); }\nwindow.speechSynthesis && window.speechSynthesis.onvoiceschanged;\n<\/script>\n\n<h2 id=\"audio-summary\" style=\"display:none;\">R\u00e9sum\u00e9 audio du chapitre<\/h2>\n<div id=\"audioText\" style=\"display:none;\">\nBienvenue dans le chapitre quantit\u00e9 de mati\u00e8re et solutions.\n\nLa quantit\u00e9 de mati\u00e8re se note n et s\u2019exprime en mole.\nUne mole contient 6 virgule 02 fois 10 puissance 23 entit\u00e9s chimiques.\nLa relation fondamentale est :\nN \u00e9gal n fois N A.\n\nPour un solide pur :\nn \u00e9gal m sur M.\nPetit m est la masse en grammes.\nGrand M est la masse molaire en grammes par mole.\n\nPour un liquide pur :\non utilise d\u2019abord la masse volumique.\nRho \u00e9gal m sur V.\nPuis :\nn \u00e9gal m sur M.\n\nPour un gaz :\nn \u00e9gal V sur V m.\nAvec V m le volume molaire.\n\u00c0 20 degr\u00e9s Celsius et pression atmosph\u00e9rique,\nV m vaut environ 24 litres par mole.\n\nPour une solution :\nla concentration molaire se note C.\nLa relation est :\nC \u00e9gal n sur V.\n\nAttention aux conversions :\nun millilitre correspond \u00e0 10 puissance moins 3 litre.\n\nPour la teneur massique :\nune solution \u00e0 6 pourcent signifie 6 grammes de solut\u00e9 pour 100 grammes de solution.\n\nAstuce importante :\nraisonner sur un litre de solution permet souvent de simplifier les calculs.\nOn utilise alors la masse volumique pour obtenir la masse totale de solution.\n\nM\u00e9thode type :\nidentifier les donn\u00e9es,\n\u00e9crire les formules,\neffectuer les conversions,\nremplacer les valeurs,\n\u00e9crire le r\u00e9sultat avec les unit\u00e9s.\n\nFin du r\u00e9sum\u00e9 du chapitre.\n<\/div>\n\n<div style=\"margin:20px 0;text-align:center;\">\n<button onclick=\"playAudioSummary()\" style=\"background:#c62828;color:white;border:none;padding:14px 24px;border-radius:10px;font-size:18px;cursor:pointer;\">\n\ud83d\udd0a \u00c9couter le r\u00e9sum\u00e9 audio du chapitre\n<\/button>\n<button onclick=\"stopAudioSummary()\" style=\"background:#333;color:white;border:none;padding:14px 24px;border-radius:10px;font-size:18px;cursor:pointer;margin-left:10px;\">\n\u23f9 Arr\u00eater\n<\/button>\n<\/div>\n\n<script>\nlet speechSynthesisUtterance;\n\nfunction playAudioSummary() {\n const text = document.getElementById(\"audioText\").innerText;\n\n speechSynthesis.cancel();\n\n speechSynthesisUtterance = new SpeechSynthesisUtterance(text);\n speechSynthesisUtterance.lang = 'fr-FR';\n speechSynthesisUtterance.rate = 0.92;\n speechSynthesisUtterance.pitch = 1;\n\n const voices = speechSynthesis.getVoices();\n const frenchVoice = voices.find(v => v.lang.includes('fr'));\n if (frenchVoice) {\n speechSynthesisUtterance.voice = frenchVoice;\n }\n\n speechSynthesis.speak(speechSynthesisUtterance);\n}\n\nfunction stopAudioSummary() {\n speechSynthesis.cancel();\n}\n<\/script>\n\n<\/body>\n<\/html>\n\n","protected":false},"excerpt":{"rendered":"<p>CH1 Chimie \u2014 Quantit\u00e9 de mati\u00e8re et solutions Premi\u00e8re sp\u00e9cialit\u00e9 physique-chimie \u00b7 Chimie CH1 \u2014 Quantit\u00e9 de mati\u00e8re et solutions Objectif : passer des grandeurs mesurables...<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-711","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages\/711","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/comments?post=711"}],"version-history":[{"count":2,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages\/711\/revisions"}],"predecessor-version":[{"id":715,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages\/711\/revisions\/715"}],"wp:attachment":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/media?parent=711"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}