{"id":245972,"date":"2025-07-06T15:29:58","date_gmt":"2025-07-06T15:29:58","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=245972"},"modified":"2025-07-06T15:30:00","modified_gmt":"2025-07-06T15:30:00","slug":"calculate-i-the-vant-hoff-factor-of-a-0-106-m-mgso4-aqueous-solution-that-has-a-freezing-point-of-0-30","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/06\/calculate-i-the-vant-hoff-factor-of-a-0-106-m-mgso4-aqueous-solution-that-has-a-freezing-point-of-0-30\/","title":{"rendered":"Calculate i, the van&#8217;t Hoff factor of a 0.106 M MgSO4 aqueous solution that has a freezing point of -0.30"},"content":{"rendered":"\n<p>Calculate i, the van&#8217;t Hoff factor of a 0.106 M MgSO4 aqueous solution that has a freezing point of -0.30 \u00c2\u00b0C. R = 0.082058 L\u00e2\u2039\u2026atm\u00e2\u2039\u2026mol\u00e2\u0081\u00bb\u00c2\u00b9\u00e2\u2039\u2026K\u00e2\u0081\u00bb\u00c2\u00b9. Kf = 1.86 \u00c2\u00b0C\/m. Report your answer to TWO places past the decimal.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To calculate the van&#8217;t Hoff factor (i), we can use the formula that relates the freezing point depression to the molality of the solution:\u0394Tf=i\u00d7Kf\u00d7m\\Delta T_f = i \\times K_f \\times m\u0394Tf\u200b=i\u00d7Kf\u200b\u00d7m<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u0394Tf\\Delta T_f\u0394Tf\u200b is the freezing point depression,<\/li>\n\n\n\n<li>iii is the van&#8217;t Hoff factor (the number of particles into which the solute dissociates),<\/li>\n\n\n\n<li>KfK_fKf\u200b is the cryoscopic constant (freezing point depression constant) of the solvent, and<\/li>\n\n\n\n<li>mmm is the molality of the solution.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Find the Freezing Point Depression (\u0394Tf\\Delta T_f\u0394Tf\u200b)<\/h3>\n\n\n\n<p>The freezing point of pure water is 0\u00b0C. Since the freezing point of the solution is -0.30\u00b0C, the freezing point depression is:\u0394Tf=0\u00b0C\u2212(\u22120.30\u00b0C)=0.30\u00b0C\\Delta T_f = 0\u00b0C &#8211; (-0.30\u00b0C) = 0.30\u00b0C\u0394Tf\u200b=0\u00b0C\u2212(\u22120.30\u00b0C)=0.30\u00b0C<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Calculate Molality (m)<\/h3>\n\n\n\n<p>Molality is defined as the moles of solute per kilogram of solvent. We are given the molarity (0.106 M) of MgSO\u2084. For dilute solutions, molarity and molality are approximately equal, so:m=0.106\u2009mol\/Lm = 0.106 \\, \\text{mol\/L}m=0.106mol\/L<\/p>\n\n\n\n<p>(Note: The density of the solution is assumed to be close to that of water for simplicity, so molarity and molality are treated as equal here.)<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Solve for the van&#8217;t Hoff Factor (i)<\/h3>\n\n\n\n<p>Now, substitute the known values into the freezing point depression equation:\u0394Tf=i\u00d7Kf\u00d7m\\Delta T_f = i \\times K_f \\times m\u0394Tf\u200b=i\u00d7Kf\u200b\u00d7m0.30=i\u00d7(1.86\u2009\u00b0C\/m)\u00d7(0.106\u2009mol\/kg)0.30 = i \\times (1.86 \\, \\text{\u00b0C\/m}) \\times (0.106 \\, \\text{mol\/kg})0.30=i\u00d7(1.86\u00b0C\/m)\u00d7(0.106mol\/kg)<\/p>\n\n\n\n<p>Solving for iii:i=0.30(1.86)(0.106)=0.300.1966\u22481.52i = \\frac{0.30}{(1.86)(0.106)} = \\frac{0.30}{0.1966} \\approx 1.52i=(1.86)(0.106)0.30\u200b=0.19660.30\u200b\u22481.52<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>The van&#8217;t Hoff factor iii is <strong>1.52<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>The van&#8217;t Hoff factor reflects the number of particles the solute dissociates into in solution. For MgSO\u2084, it dissociates into two ions: Mg\u00b2\u207a and SO\u2084\u00b2\u207b. Ideally, this would give an iii of 2. However, due to ion pairing or slight deviations in ideal dissociation, the experimental value of iii here is 1.52, which suggests that some degree of ion pairing or incomplete dissociation occurs in this specific solution.<\/p>\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-banner5-760.jpeg\" alt=\"\" class=\"wp-image-245973\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Calculate i, the van&#8217;t Hoff factor of a 0.106 M MgSO4 aqueous solution that has a freezing point of -0.30 \u00c2\u00b0C. R = 0.082058 L\u00e2\u2039\u2026atm\u00e2\u2039\u2026mol\u00e2\u0081\u00bb\u00c2\u00b9\u00e2\u2039\u2026K\u00e2\u0081\u00bb\u00c2\u00b9. Kf = 1.86 \u00c2\u00b0C\/m. Report your answer to TWO places past the decimal. The Correct Answer and Explanation is: To calculate the van&#8217;t Hoff factor (i), we can use the [&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-245972","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245972","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=245972"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245972\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=245972"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=245972"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=245972"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}