{"id":185906,"date":"2025-01-23T17:30:54","date_gmt":"2025-01-23T17:30:54","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=185906"},"modified":"2025-01-23T17:30:56","modified_gmt":"2025-01-23T17:30:56","slug":"one-gram-of-sucrose-molecular-weight-342-is-dissolved-in-100-ml-of-solution-at-25c","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/23\/one-gram-of-sucrose-molecular-weight-342-is-dissolved-in-100-ml-of-solution-at-25c\/","title":{"rendered":"One gram of sucrose, molecular weight 342, is dissolved in 100 mL of solution at 25\u00b0C"},"content":{"rendered":"\n<p>One gram of sucrose, molecular weight 342, is dissolved in 100 mL of solution at 25\u00b0C. What is the osmotic pressure of the solution?<\/p>\n\n\n\n<p>Select one<\/p>\n\n\n\n<p>a 0.082 atm<\/p>\n\n\n\n<p>b. 005 atm<\/p>\n\n\n\n<p>C.002 atm<\/p>\n\n\n\n<p>d. 0.0071 atm<\/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 osmotic pressure of a sucrose solution, we can use the formula for osmotic pressure: \u03a0=nVRT\\Pi = \\frac{n}{V}RT<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u03a0\\Pi is the osmotic pressure (in atm),<\/li>\n\n\n\n<li>nn is the number of moles of solute,<\/li>\n\n\n\n<li>VV is the volume of the solution in liters,<\/li>\n\n\n\n<li>RR is the ideal gas constant (0.0821 L\u00b7atm\/mol\u00b7K),<\/li>\n\n\n\n<li>TT is the temperature in Kelvin (25\u00b0C = 298.15 K).<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate the number of moles of sucrose<\/h3>\n\n\n\n<p>First, we need to determine the number of moles of sucrose dissolved in the solution. The molecular weight of sucrose (C\u2081\u2082H\u2082\u2082O\u2081\u2081) is given as 342 g\/mol.<\/p>\n\n\n\n<p>The mass of sucrose is 1 gram, so the number of moles of sucrose is: moles&nbsp;of&nbsp;sucrose=mass&nbsp;of&nbsp;sucrosemolecular&nbsp;weight=1\u2009g342\u2009g\/mol=0.00292\u2009mol\\text{moles of sucrose} = \\frac{\\text{mass of sucrose}}{\\text{molecular weight}} = \\frac{1 \\, \\text{g}}{342 \\, \\text{g\/mol}} = 0.00292 \\, \\text{mol}<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Convert the volume to liters<\/h3>\n\n\n\n<p>The volume of the solution is given as 100 mL, which is 0.100 L.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Plug the values into the osmotic pressure formula<\/h3>\n\n\n\n<p>Now we can calculate the osmotic pressure: \u03a0=0.00292\u2009mol0.100\u2009L\u00d70.0821\u2009L\\cdotpatm\/mol\\cdotpK\u00d7298.15\u2009K\\Pi = \\frac{0.00292 \\, \\text{mol}}{0.100 \\, \\text{L}} \\times 0.0821 \\, \\text{L\u00b7atm\/mol\u00b7K} \\times 298.15 \\, \\text{K} \u03a0=0.002920.100\u00d70.0821\u00d7298.15=0.0071\u2009atm\\Pi = \\frac{0.00292}{0.100} \\times 0.0821 \\times 298.15 = 0.0071 \\, \\text{atm}<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Select the correct answer<\/h3>\n\n\n\n<p>From the calculated value, we find that the osmotic pressure is approximately 0.0071 atm, which corresponds to <strong>option d<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation of the Osmotic Pressure<\/h3>\n\n\n\n<p>Osmotic pressure is the pressure exerted by a solvent when it passes through a semipermeable membrane to balance the concentration of solute on both sides. For dilute solutions like the one we are dealing with here, osmotic pressure is directly proportional to the solute concentration and temperature, and this is why it increases as either the solute concentration or temperature increases. Since sucrose is a non-volatile solute and doesn&#8217;t ionize in solution, its effect on osmotic pressure is straightforward, with the number of moles of sucrose determining the pressure exerted on the solvent.<\/p>\n\n\n\n<p>Therefore, the osmotic pressure of the solution is <strong>0.0071 atm<\/strong>, and the correct answer is <strong>d. 0.0071 atm<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>One gram of sucrose, molecular weight 342, is dissolved in 100 mL of solution at 25\u00b0C. What is the osmotic pressure of the solution? Select one a 0.082 atm b. 005 atm C.002 atm d. 0.0071 atm The Correct Answer and Explanation is : To calculate the osmotic pressure of a sucrose solution, we can [&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-185906","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185906","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=185906"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185906\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=185906"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=185906"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=185906"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}