{"id":239513,"date":"2025-07-02T21:40:54","date_gmt":"2025-07-02T21:40:54","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=239513"},"modified":"2025-07-02T21:40:56","modified_gmt":"2025-07-02T21:40:56","slug":"assume-that-carbon-dioxide-in-a-2-00-l-soda-bottle-is-dissolved-at-a-pressure-of-2-07-bar","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/02\/assume-that-carbon-dioxide-in-a-2-00-l-soda-bottle-is-dissolved-at-a-pressure-of-2-07-bar\/","title":{"rendered":"Assume that carbon dioxide in a 2.00 L soda bottle is dissolved at a pressure of 2.07 bar."},"content":{"rendered":"\n<p>Assume that carbon dioxide in a 2.00 L soda bottle is dissolved at a pressure of 2.07 bar. At 2.28\u00c2\u00b0C, the Henry&#8217;s law constant for carbon dioxide dissolved in water is 0.068 L\/bar. Calculate the concentration of dissolved carbon dioxide in mol\/L and the mass of carbon dioxide that can be dissolved in the bottle of soda at the given pressure.<\/p>\n\n\n\n<p><strong>The Correct Answer and Explanation is:<\/strong><sup data-fn=\"20450495-d392-4bb9-b862-d83ea760f6f0\" class=\"fn\"><a id=\"20450495-d392-4bb9-b862-d83ea760f6f0-link\" href=\"#20450495-d392-4bb9-b862-d83ea760f6f0\">1<\/a><\/sup><\/p>\n\n\n\n\n\n<p>Here is a step-by-step solution to the problem presented in the image.<\/p>\n\n\n\n<p><strong>Given Information:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume of soda (V) = 2.00 L<\/li>\n\n\n\n<li>Pressure of CO\u2082 (P_CO\u2082) = 2.07 bar<\/li>\n\n\n\n<li>Henry&#8217;s Law constant (k_H) = 0.068 mol \/ (L\u00b7bar)<\/li>\n\n\n\n<li>The problem asks for the concentration (in mol\/L) and the mass (in grams) of the dissolved carbon dioxide.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Part 1: Calculate the concentration of dissolved CO\u2082<\/strong><\/h3>\n\n\n\n<p>We will provided.<\/p>\n\n\n\n<p>This problem uses Henry&#8217;s Law, which states that the concentration of a dissolved gas in a liquid is directly proportional to the partial pressure of that gas above the liquid.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Given Information:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Volume of soda (V):<\/strong>\u00a02.00 L<\/li>\n\n\n\n<li><strong>Pressure of CO\u2082 (P):<\/strong>\u00a02.07 bar<\/li>\n\n\n\n<li><strong>Henry&#8217;s Law Constant (k_H):<\/strong>\u00a00.068 mol \/ (L\u00b7bar)<\/li>\n\n\n\n<li><strong>Molar Mass of CO\u2082 (M):<\/strong>\n<ul class=\"wp-block-list\">\n<li>C = 12.01\u00a0g\/mol<\/li>\n\n\n\n<li>O = 16.00\u00a0g\/mol<\/li>\n\n\n\n<li>M(CO\u2082) = 12.01 + 2(16.00) = 44.01\u00a0g\/mol<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Part 1: Calculate the Concentration of Dissolved CO\u2082<\/strong><\/h3>\n\n\n\n<p>The formula for Henry&#8217;s Law is:<br><strong>C = k_H \u00d7 P<\/strong><\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>C<\/strong>\u00a0is the concentration in mol\/L<\/li>\n\n\n\n<li><strong>k_H<\/strong>\u00a0is the Henry&#8217;s Law constant<\/li>\n\n\n\n<li><strong>P<\/strong>\u00a0is the partial pressure of the gas<\/li>\n<\/ul>\n\n\n\n<p><strong>Calculation:<\/strong><br>C = (0.068 mol \/ (L\u00b7bar)) \u00d7 (2.07 bar)<br>C = 0.14076 mol\/L<\/p>\n\n\n\n<p>Rounding to the correct number of significant figures (2, limited by the constant 0.068):<br><strong>C \u2248 0.14 mol\/L<\/strong><\/p>\n\n\n\n<p><strong>The concentration of dissolved carbon dioxide is 0.14 mol\/L.<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Part 2: Calculate the Mass of Dissolved CO\u2082<\/strong><\/h3>\n\n\n\n<p>First, we need to find the total number of moles (n) of CO\u2082 dissolved in the 2.00 L bottle using the concentration we just calculated.<\/p>\n\n\n\n<p><strong>Formula:<\/strong><br><strong>n = C \u00d7 V<\/strong><\/p>\n\n\n\n<p><strong>Calculation (using the unrounded concentration for accuracy):<\/strong><br>n = (0.14076 mol\/L) \u00d7 (2.00 L)<br>n = 0.28152 mol<\/p>\n\n\n\n<p>Next use Henry&#8217;s Law, which states that the concentration (C) of a dissolved gas is directly proportional to the partial pressure (P) of that gas above the solution.<\/p>\n\n\n\n<p><strong>Formula:<\/strong><br>C = k_H \u00d7 P<\/p>\n\n\n\n<p><strong>Calculation:<\/strong><br>Substitute the given values into the formula:<br>C = (0.068 mol \/ (L\u00b7bar)) \u00d7 (2.07 bar)<br>C = 0.14076 mol\/L<\/p>\n\n\n\n<p><strong>Significant Figures:<\/strong><br>The Henry&#8217;s law constant (0.068) has two significant figures, and the pressure (2.07) has three. The result should be rounded to the least number of significant figures, which is two.<\/p>\n\n\n\n<p><strong>Concentration = 0.14 mol\/L<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Part 2: Calculate the mass of dissolved CO\u2082<\/strong><\/h3>\n\n\n\n<p>First, we need to find the total number of moles (n) of CO\u2082 dissolved in the 2.00 L bottle. Then, we can convert moles to mass using the molar mass of CO\u2082.<\/p>\n\n\n\n<p><strong>Step 2a: Calculate the moles of CO\u2082<\/strong><br><strong>Formula:<\/strong><br>Moles (n) = Concentration (C) \u00d7 Volume (, we convert the moles of CO\u2082 to mass (m) using its molar mass.<\/p>\n\n\n\n<p><strong>Formula:<\/strong><br><strong>m = n \u00d7 M<\/strong><\/p>\n\n\n\n<p><strong>Calculation:<\/strong><br>m = (0.28152 mol) \u00d7 (44.01&nbsp;g\/mol&nbsp;)<br>m = 12.3896 g<\/p>\n\n\n\n<p>Rounding to the correct number of significant figures (2, again limited by the constant 0.068):<br><strong>m \u2248 12 g<\/strong><\/p>\n\n\n\n<p>**The mass of carbon dioxide that can be dissolved in the bottle is 12 g.**V)<\/p>\n\n\n\n<p><strong>Calculation:<\/strong><br>Using the unrounded concentration from the previous step to maintain accuracy:<br>n = (0.14076 mol\/L) \u00d7 (2.00 L)<br>n = 0.28152 mol<\/p>\n\n\n\n<p><strong>Step 2b: Calculate the mass of CO\u2082<\/strong><br><strong>Formula:<\/strong><br>Mass (m) = Moles (n) \u00d7 Molar Mass (M)<\/p>\n\n\n\n<p><strong>Molar Mass of CO\u2082:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon (C) \u2248 12.01\u00a0g\/mol<\/li>\n\n\n\n<li>Oxygen (O) \u2248 16.00\u00a0g\/mol<\/li>\n\n\n\n<li>M_CO\u2082 = 12.01 + 2(16.00) = 44.01\u00a0g\/mol<\/li>\n<\/ul>\n\n\n\n<p><strong>Calculation:<\/strong><br>m = (0.28152 mol) \u00d7 (44.01&nbsp;g\/mol&nbsp;)<br>m = 12.3896 g<\/p>\n\n\n\n<p><strong>Significant Figures:<\/strong><br>The calculation involves values with two (from k_H), three (from V), and three (from P) significant figures. The final answer must be rounded to two significant figures.<\/p>\n\n\n\n<p><strong>Mass = 12 g<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Final Answer:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The concentration of dissolved carbon dioxide is\u00a0<strong>0.14 mol\/L<\/strong>.<\/li>\n\n\n\n<li>The mass of carbon dioxide that can be dissolved in the bottle is\u00a0<strong>12 g<\/strong>.<\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-40.jpeg\" alt=\"\" class=\"wp-image-239515\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Assume that carbon dioxide in a 2.00 L soda bottle is dissolved at a pressure of 2.07 bar. At 2.28\u00c2\u00b0C, the Henry&#8217;s law constant for carbon dioxide dissolved in water is 0.068 L\/bar. Calculate the concentration of dissolved carbon dioxide in mol\/L and the mass of carbon dioxide that can be dissolved in the bottle [&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-239513","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/239513","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=239513"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/239513\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=239513"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=239513"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=239513"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}