{"id":256802,"date":"2025-07-17T03:16:59","date_gmt":"2025-07-17T03:16:59","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=256802"},"modified":"2025-07-17T03:17:01","modified_gmt":"2025-07-17T03:17:01","slug":"kf-and-kb-for-benzene-are-5-07-k-kg-mol-and-2-53-k-kg-mol-respectively-3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/17\/kf-and-kb-for-benzene-are-5-07-k-kg-mol-and-2-53-k-kg-mol-respectively-3\/","title":{"rendered":"Kf and Kb for benzene are 5.07 K kg\/mol and 2.53 K kg\/mol respectively."},"content":{"rendered":"\n

Kf and Kb for benzene are 5.07 K kg\/mol and 2.53 K kg\/mol respectively. A solution of a non-volatile solute in benzene showed an elevation of 0.50 K. What should be the depression of the freezing point of this solution?<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To solve this problem, we can use the relationship between the elevation in boiling point (\u0394T_b) and the depression in freezing point (\u0394T_f), which are related by the following formulas:<\/p>\n\n\n\n

1. Boiling Point Elevation Formula:<\/strong><\/h3>\n\n\n\n

\u0394Tb=Kb\u22c5m\\Delta T_b = K_b \\cdot m\u0394Tb\u200b=Kb\u200b\u22c5m<\/p>\n\n\n\n

Where:<\/p>\n\n\n\n

    \n
  • \u0394Tb\\Delta T_b\u0394Tb\u200b is the elevation in boiling point.<\/li>\n\n\n\n
  • KbK_bKb\u200b is the ebullioscopic constant (for benzene, Kb=2.53\u2009K\u00a0kg\/molK_b = 2.53 \\, \\text{K kg\/mol}Kb\u200b=2.53K\u00a0kg\/mol).<\/li>\n\n\n\n
  • mmm is the molality of the solution.<\/li>\n<\/ul>\n\n\n\n

    2. Freezing Point Depression Formula:<\/strong><\/h3>\n\n\n\n

    \u0394Tf=Kf\u22c5m\\Delta T_f = K_f \\cdot m\u0394Tf\u200b=Kf\u200b\u22c5m<\/p>\n\n\n\n

    Where:<\/p>\n\n\n\n

      \n
    • \u0394Tf\\Delta T_f\u0394Tf\u200b is the depression in freezing point.<\/li>\n\n\n\n
    • KfK_fKf\u200b is the cryoscopic constant (for benzene, Kf=5.07\u2009K\u00a0kg\/molK_f = 5.07 \\, \\text{K kg\/mol}Kf\u200b=5.07K\u00a0kg\/mol).<\/li>\n\n\n\n
    • mmm is the molality of the solution.<\/li>\n<\/ul>\n\n\n\n

      Step-by-step Solution:<\/h3>\n\n\n\n

      Step 1: Find the Molality (m) of the Solution<\/h4>\n\n\n\n

      The elevation of the boiling point is given as \u0394Tb=0.50\u2009K\\Delta T_b = 0.50 \\, \\text{K}\u0394Tb\u200b=0.50K.<\/p>\n\n\n\n

      Using the formula for boiling point elevation:\u0394Tb=Kb\u22c5m\\Delta T_b = K_b \\cdot m\u0394Tb\u200b=Kb\u200b\u22c5m<\/p>\n\n\n\n

      Substitute the known values:0.50=2.53\u22c5m0.50 = 2.53 \\cdot m0.50=2.53\u22c5m<\/p>\n\n\n\n

      Solving for mmm:m=0.502.53=0.197\u2009mol\/kgm = \\frac{0.50}{2.53} = 0.197 \\, \\text{mol\/kg}m=2.530.50\u200b=0.197mol\/kg<\/p>\n\n\n\n

      Step 2: Calculate the Freezing Point Depression (\u0394T_f)<\/h4>\n\n\n\n

      Now that we know the molality m=0.197\u2009mol\/kgm = 0.197 \\, \\text{mol\/kg}m=0.197mol\/kg, we can use the freezing point depression formula:\u0394Tf=Kf\u22c5m\\Delta T_f = K_f \\cdot m\u0394Tf\u200b=Kf\u200b\u22c5m<\/p>\n\n\n\n

      Substitute the values for KfK_fKf\u200b and mmm:\u0394Tf=5.07\u22c50.197\\Delta T_f = 5.07 \\cdot 0.197\u0394Tf\u200b=5.07\u22c50.197\u0394Tf=1.00\u2009K\\Delta T_f = 1.00 \\, \\text{K}\u0394Tf\u200b=1.00K<\/p>\n\n\n\n

      Final Answer:<\/h3>\n\n\n\n

      The depression in freezing point (\u0394Tf\\Delta T_f\u0394Tf\u200b) is 1.00 K<\/strong>.<\/p>\n\n\n\n

      Explanation:<\/h3>\n\n\n\n

      In this problem, we first used the given boiling point elevation to calculate the molality of the solution. This molality was then applied to the freezing point depression formula to find the change in the freezing point.<\/p>\n\n\n\n

      Boiling point elevation and freezing point depression are colligative properties, meaning they depend only on the concentration of solute particles, not on their identity. The solution behaves similarly for both boiling and freezing point changes, though the constants KbK_bKb\u200b and KfK_fKf\u200b differ for each property.<\/p>\n\n\n\n

      In this case, the solute causes an increase in the boiling point and a decrease in the freezing point, with the respective changes calculated based on the molality of the solution. The relationship between boiling point elevation and freezing point depression is crucial when understanding the behavior of solutions in terms of phase transitions.<\/p>\n\n\n\n

      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

      Kf and Kb for benzene are 5.07 K kg\/mol and 2.53 K kg\/mol respectively. A solution of a non-volatile solute in benzene showed an elevation of 0.50 K. What should be the depression of the freezing point of this solution? The Correct Answer and Explanation is: To solve this problem, we can use the relationship […]<\/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-256802","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/256802","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=256802"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/256802\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=256802"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=256802"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=256802"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}