In the gaseous state chlorine exists as a diatomic molecule Cl2 (molar mass=70.9g\/mol). Calculate the number of moles present in 120 g of chlorine gas.<\/p>\n\n\n\n
Express the quantity and moles to three significant figures.<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the number of moles<\/strong> of chlorine gas (Cl\u2082) in 120 grams<\/strong>, use the following formula:<\/p>\n\n\n\n [ Given:<\/p>\n\n\n\n [ Answer:<\/strong> The concept of mole<\/strong> is fundamental in chemistry and allows chemists to count particles (atoms, molecules, ions) by weighing them. One mole of any substance contains Avogadro\u2019s number<\/strong> of particles, which is approximately ( 6.022 \\times 10^{23} ) entities.<\/p>\n\n\n\n In this problem, we are given chlorine gas<\/strong>, which naturally exists as diatomic molecules<\/strong> (Cl\u2082), not individual chlorine atoms. This is important because it affects the molar mass<\/strong> used in the calculation. While the atomic mass of a single chlorine atom is about 35.45 g\/mol<\/strong>, the molar mass of Cl\u2082 is:<\/p>\n\n\n\n [ This value represents the mass of one mole of Cl\u2082 molecules.<\/p>\n\n\n\n To determine how many moles are in 120 grams of chlorine gas, we divide the total mass by the molar mass. The formula used is:<\/p>\n\n\n\n [ Substituting the values:<\/p>\n\n\n\n [ Since the question specifies to express the answer to three significant figures<\/strong>, we round 1.692<\/strong> to 1.69 mol<\/strong>.<\/p>\n\n\n\n This value tells us that 120 grams of chlorine gas contains approximately 1.69 moles<\/strong>, or equivalently, about ( 1.69 \\times 6.022 \\times 10^{23} \\approx 1.02 \\times 10^{24} ) chlorine molecules.<\/p>\n\n\n\n Understanding molar conversions like this is essential in stoichiometry, allowing scientists to relate mass to number of particles in chemical reactions, gas laws, and concentration calculations.<\/p>\n","protected":false},"excerpt":{"rendered":" In the gaseous state chlorine exists as a diatomic molecule Cl2 (molar mass=70.9g\/mol). Calculate the number of moles present in 120 g of chlorine gas. Express the quantity and moles to three significant figures. The correct answer and explanation is : To calculate the number of moles of chlorine gas (Cl\u2082) in 120 grams, use […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-210702","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/210702","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=210702"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/210702\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=210702"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=210702"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=210702"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\\text{Moles} = \\frac{\\text{Mass (g)}}{\\text{Molar Mass (g\/mol)}}
]<\/p>\n\n\n\n
\n\n\n\nStep-by-step Calculation:<\/strong><\/h3>\n\n\n\n
\n
\\text{Moles} = \\frac{120\\ \\text{g}}{70.9\\ \\text{g\/mol}} \\approx 1.692\\ \\text{mol}
]<\/p>\n\n\n\n
[
\\boxed{1.69\\ \\text{mol}}
]<\/p>\n\n\n\n
\n\n\n\nExplanation (300+ words):<\/strong><\/h3>\n\n\n\n
35.45 \\times 2 = 70.9\\ \\text{g\/mol}
]<\/p>\n\n\n\n
\\text{Moles} = \\frac{\\text{Mass}}{\\text{Molar Mass}}
]<\/p>\n\n\n\n
\\frac{120}{70.9} \\approx 1.692\\ \\text{mol}
]<\/p>\n\n\n\n