{"id":196528,"date":"2025-03-05T05:32:12","date_gmt":"2025-03-05T05:32:12","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=196528"},"modified":"2025-03-05T05:32:15","modified_gmt":"2025-03-05T05:32:15","slug":"what-is-the-density-of-co2-gas-at-stp-conditions-if-2-50-g-occupies-5-60-l-at-789-torr","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/05\/what-is-the-density-of-co2-gas-at-stp-conditions-if-2-50-g-occupies-5-60-l-at-789-torr\/","title":{"rendered":"What is the density of CO2 gas at STP conditions if 2.50 g occupies 5.60 L at 789 torr"},"content":{"rendered":"\n
What is the density of CO2 gas at STP conditions if 2.50 g occupies 5.60 L at 789 torr?<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n
To find the density of CO\u2082 gas under standard temperature and pressure (STP) conditions, we can use the following steps.<\/p>\n\n\n\n
Given:<\/h3>\n\n\n\n
\n
Mass of CO\u2082 = 2.50 g<\/li>\n\n\n\n
Volume of CO\u2082 = 5.60 L<\/li>\n\n\n\n
Pressure = 789 torr<\/li>\n\n\n\n
Standard Temperature = 273.15 K (STP)<\/li>\n\n\n\n
Standard Pressure = 1 atm = 760 torr<\/li>\n<\/ul>\n\n\n\n
Step 1: Use the Ideal Gas Law to Find the Moles of CO\u2082<\/h3>\n\n\n\n
The ideal gas law equation is: [ PV = nRT ] Where:<\/p>\n\n\n\n
\n
( P ) is the pressure<\/li>\n\n\n\n
( V ) is the volume<\/li>\n\n\n\n
( n ) is the number of moles of gas<\/li>\n\n\n\n
( R ) is the ideal gas constant (0.0821 L\u00b7atm\/(mol\u00b7K))<\/li>\n\n\n\n
( T ) is the temperature<\/li>\n<\/ul>\n\n\n\n
We will first need to convert the pressure to atm and use the given volume and temperature to find the number of moles of CO\u2082 at the given conditions.<\/p>\n\n\n\n
\n
Convert pressure from torr to atm: [ \\frac{789 \\, \\text{torr}}{760 \\, \\text{torr\/atm}} = 1.038 \\, \\text{atm} ]<\/li>\n<\/ul>\n\n\n\n
Now, apply the ideal gas law: [ P = 1.038 \\, \\text{atm}, \\quad V = 5.60 \\, \\text{L}, \\quad T = 273.15 \\, \\text{K} ] Rearrange the ideal gas law to solve for ( n ) (moles of CO\u2082): [ n = \\frac{PV}{RT} ] Substituting the values: [ n = \\frac{(1.038 \\, \\text{atm})(5.60 \\, \\text{L})}{(0.0821 \\, \\text{L\u00b7atm\/mol\u00b7K})(273.15 \\, \\text{K})} ] [ n = 0.247 \\, \\text{mol} ]<\/p>\n\n\n\n
Step 2: Calculate the Molar Mass of CO\u2082<\/h3>\n\n\n\n
Now that we know the number of moles of CO\u2082, we can calculate the molar mass. The molar mass of CO\u2082 is: [ M_{\\text{CO}2} = \\frac{\\text{mass}}{\\text{moles}} ] [ M<\/em>{\\text{CO}_2} = \\frac{2.50 \\, \\text{g}}{0.247 \\, \\text{mol}} = 10.1 \\, \\text{g\/mol} ]<\/p>\n\n\n\n
Step 3: Calculate the Density of CO\u2082<\/h3>\n\n\n\n
The density of a gas is given by: [ \\text{Density} = \\frac{\\text{Mass}}{\\text{Volume}} ] Substituting the values: [ \\text{Density} = \\frac{2.50 \\, \\text{g}}{5.60 \\, \\text{L}} = 0.446 \\, \\text{g\/L} ]<\/p>\n\n\n\n
Conclusion:<\/h3>\n\n\n\n
The density of CO\u2082 gas under the given conditions (789 torr and 5.60 L volume) is 0.446 g\/L<\/strong>.<\/p>\n\n\n\n
Let me also provide an image of the calculation steps:<\/p>\n\n\n\n<\/figure>\n\n\n\n
Here is the detailed calculation sheet showing the steps to find the density of CO\u2082 gas under the given conditions. It includes the ideal gas law equation, conversions, and calculations for moles, molar mass, and density. <\/p>\n","protected":false},"excerpt":{"rendered":"
What is the density of CO2 gas at STP conditions if 2.50 g occupies 5.60 L at 789 torr? The Correct Answer and Explanation is : To find the density of CO\u2082 gas under standard temperature and pressure (STP) conditions, we can use the following steps. Given: Step 1: Use the Ideal Gas Law to […]<\/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-196528","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196528","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=196528"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196528\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=196528"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=196528"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=196528"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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