{"id":213989,"date":"2025-05-13T09:05:20","date_gmt":"2025-05-13T09:05:20","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=213989"},"modified":"2025-05-13T09:05:22","modified_gmt":"2025-05-13T09:05:22","slug":"freezing-point-for-t-butanol-is-25-50-degrees-c-and-kf-is-9-1-degrees-c-m","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/13\/freezing-point-for-t-butanol-is-25-50-degrees-c-and-kf-is-9-1-degrees-c-m\/","title":{"rendered":"freezing point for t-butanol is 25.50 degrees C and kf is 9.1 degrees C\/m."},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">The freezing point for t-butanol is 25.50 degrees C and kf is 9.1 degrees C\/m. Usually t-butanol absorbs water on exposure to the air. If the freezing point of a 10.0 g sample of t-butanol is measured as 24.59 degrees C, how many grams of water are present in the sample?<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">To determine how many grams of water are present in the t-butanol sample, we need to apply the freezing point depression formula. Freezing point depression occurs when a solute (in this case, water) is dissolved in a solvent (t-butanol). The formula for freezing point depression is: \u0394Tf=Kf\u00d7m\\Delta T_f = K_f \\times m<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u0394Tf\\Delta T_f is the freezing point depression,<\/li>\n\n\n\n<li>KfK_f is the cryoscopic constant of the solvent (in this case, t-butanol),<\/li>\n\n\n\n<li>mm is the molality of the solution.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate the freezing point depression (\u0394Tf\\Delta T_f)<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The freezing point of pure t-butanol is 25.50\u00b0C, and the measured freezing point of the sample with water is 24.59\u00b0C. Therefore, the freezing point depression is: \u0394Tf=25.50\u2218C\u221224.59\u2218C=0.91\u2218C\\Delta T_f = 25.50^\\circ C &#8211; 24.59^\\circ C = 0.91^\\circ C<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Use the freezing point depression formula<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Now that we know the freezing point depression (\u0394Tf=0.91\u2218C\\Delta T_f = 0.91^\\circ C) and the cryoscopic constant (Kf=9.1\u2218C\/mK_f = 9.1^\\circ C\/m), we can use the formula to solve for the molality (mm) of the solution: m=\u0394TfKf=0.91\u2218C9.1\u2218C\/m=0.1\u2009mol\/kgm = \\frac{\\Delta T_f}{K_f} = \\frac{0.91^\\circ C}{9.1^\\circ C\/m} = 0.1 \\, \\text{mol\/kg}<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Determine the amount of water in the sample<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The molality (mm) is defined as the number of moles of solute (water) per kilogram of solvent (t-butanol). Since we know the sample of t-butanol weighs 10.0 g, or 0.0100 kg, we can calculate the moles of water: m=moles&nbsp;of&nbsp;watermass&nbsp;of&nbsp;t-butanol&nbsp;in&nbsp;kgm = \\frac{\\text{moles of water}}{\\text{mass of t-butanol in kg}}<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Rearranging for moles of water: moles&nbsp;of&nbsp;water=m\u00d7mass&nbsp;of&nbsp;t-butanol&nbsp;in&nbsp;kg=0.1\u2009mol\/kg\u00d70.0100\u2009kg=0.001\u2009mol\\text{moles of water} = m \\times \\text{mass of t-butanol in kg} = 0.1 \\, \\text{mol\/kg} \\times 0.0100 \\, \\text{kg} = 0.001 \\, \\text{mol}<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Now, we can convert moles of water to grams. The molar mass of water is 18.015 g\/mol: grams&nbsp;of&nbsp;water=moles&nbsp;of&nbsp;water\u00d7molar&nbsp;mass&nbsp;of&nbsp;water=0.001\u2009mol\u00d718.015\u2009g\/mol=0.018\u2009g\\text{grams of water} = \\text{moles of water} \\times \\text{molar mass of water} = 0.001 \\, \\text{mol} \\times 18.015 \\, \\text{g\/mol} = 0.018 \\, \\text{g}<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The sample of t-butanol contains <strong>0.018 grams<\/strong> of water.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Freezing point depression is a colligative property, meaning it depends on the number of solute particles, not their identity. When water dissolves in t-butanol, it lowers the freezing point, which is the principle behind this calculation. By using the freezing point depression formula and knowing the mass of t-butanol and the cryoscopic constant, we can calculate the amount of water present based on the observed change in freezing point.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The freezing point for t-butanol is 25.50 degrees C and kf is 9.1 degrees C\/m. Usually t-butanol absorbs water on exposure to the air. If the freezing point of a 10.0 g sample of t-butanol is measured as 24.59 degrees C, how many grams of water are present in the sample? The Correct Answer and [&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-213989","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/213989","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=213989"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/213989\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=213989"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=213989"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=213989"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}