{"id":196572,"date":"2025-03-05T13:16:00","date_gmt":"2025-03-05T13:16:00","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=196572"},"modified":"2025-03-05T13:16:03","modified_gmt":"2025-03-05T13:16:03","slug":"what-is-the-approximate-density-of-1-mole-of-fluorine-gas-f2-molar-mass-70-1-g-mol-in-units-of-grams-per-liter-at-stp-given-that-1-mol-22-414-l-for-an-ideal-gas-at-stpa-0-6b-4-5c-3-9d-1-7","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/05\/what-is-the-approximate-density-of-1-mole-of-fluorine-gas-f2-molar-mass-70-1-g-mol-in-units-of-grams-per-liter-at-stp-given-that-1-mol-22-414-l-for-an-ideal-gas-at-stpa-0-6b-4-5c-3-9d-1-7\/","title":{"rendered":"What is the approximate density of 1 mole of fluorine gas, F2, (molar mass = 70.1 g\/mol) in units of grams per liter at STP, given that 1 mol = 22.414 L for an ideal gas at STP?A 0.6B 4.5C 3.9D 1.7"},"content":{"rendered":"\n<p>What is the approximate density of 1 mole of fluorine gas, F2, (molar mass = 70.1 g\/mol) in units of grams per liter at STP, given that 1 mol = 22.414 L for an ideal gas at STP?<br>A 0.6<br>B 4.5<br>C 3.9<br>D 1.7<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>To calculate the approximate density of fluorine gas ((\\text{F}_2)) at standard temperature and pressure (STP), we use the density formula for gases:<\/p>\n\n\n\n<p>[<br>\\text{Density} = \\frac{\\text{Molar Mass}}{\\text{Molar Volume}}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Given data:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Molar mass of (\\text{F}_2)<\/strong> = 70.1 g\/mol<\/li>\n\n\n\n<li><strong>Molar volume at STP<\/strong> = 22.414 L\/mol<\/li>\n<\/ul>\n\n\n\n<p>Substituting the values:<\/p>\n\n\n\n<p>[<br>\\text{Density} = \\frac{70.1 \\text{ g\/mol}}{22.414 \\text{ L\/mol}}<br>]<\/p>\n\n\n\n<p>[<br>\\approx 3.13 \\text{ g\/L}<br>]<\/p>\n\n\n\n<p>Since none of the answer choices list <strong>3.13 g\/L<\/strong>, we look for the closest option. The closest value is <strong>3.9 g\/L<\/strong>, meaning the correct answer is <strong>C: 3.9 g\/L<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>Density is defined as mass per unit volume. For gases at STP, we use the fact that <strong>1 mole of any ideal gas occupies 22.414 liters<\/strong>. By dividing the molar mass of fluorine gas (which consists of two fluorine atoms, hence <strong>70.1 g\/mol<\/strong>) by the molar volume, we obtain its density.<\/p>\n\n\n\n<p>Although the exact calculated value is <strong>3.13 g\/L<\/strong>, it&#8217;s possible that rounding in the provided answer choices led to <strong>3.9 g\/L<\/strong> being the closest available answer.<\/p>\n\n\n\n<p>This method applies to any gas where its molar mass is known. The higher the molar mass, the denser the gas will be under the same conditions.<\/p>\n\n\n\n<p>Let me generate an image illustrating this concept.<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/03\/image-229-1024x777.png\" alt=\"\" class=\"wp-image-196573\"\/><\/figure>\n\n\n\n<p>Here is a visual representation of the density of fluorine gas ((\\text{F}_2)) at STP. The calculated density is approximately <strong>3.13 g\/L<\/strong>, and the closest answer choice provided is <strong>3.9 g\/L (C)<\/strong>. <\/p>\n","protected":false},"excerpt":{"rendered":"<p>What is the approximate density of 1 mole of fluorine gas, F2, (molar mass = 70.1 g\/mol) in units of grams per liter at STP, given that 1 mol = 22.414 L for an ideal gas at STP?A 0.6B 4.5C 3.9D 1.7 The correct answer and explanation is : To calculate the approximate density of [&hellip;]<\/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-196572","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196572","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=196572"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196572\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=196572"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=196572"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=196572"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}