{"id":194601,"date":"2025-02-24T09:51:22","date_gmt":"2025-02-24T09:51:22","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=194601"},"modified":"2025-02-24T09:51:25","modified_gmt":"2025-02-24T09:51:25","slug":"isopropyl-alcohol-70-ethyl-alcohol-70-calculate-the-freezing-point-depression","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/24\/isopropyl-alcohol-70-ethyl-alcohol-70-calculate-the-freezing-point-depression\/","title":{"rendered":"Isopropyl Alcohol 70% Ethyl Alcohol 70% Calculate the freezing point depression"},"content":{"rendered":"\n<p>Isopropyl Alcohol 70% Ethyl Alcohol 70% Calculate the freezing point depression<\/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 <strong>freezing point depression<\/strong> ((\\Delta T_f)) for <strong>70% isopropyl alcohol<\/strong> and <strong>70% ethyl alcohol<\/strong>, we use the <strong>freezing point depression equation<\/strong>:<\/p>\n\n\n\n<p>[<br>\\Delta T_f = i \\cdot K_f \\cdot m<br>]<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( i ) = van&#8217;t Hoff factor (for non-electrolytes like alcohols, ( i = 1 ))<\/li>\n\n\n\n<li>( K_f ) = cryoscopic constant (for water, ( K_f = 1.86 \\,^{\\circ}C \\cdot \\text{kg\/mol} ))<\/li>\n\n\n\n<li>( m ) = molality (moles of solute per kg of solvent)<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 1: Define Given Data<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>70% isopropyl alcohol (C\u2083H\u2087OH)<\/strong> and <strong>70% ethyl alcohol (C\u2082H\u2085OH)<\/strong> indicate <strong>70 g of alcohol per 100 g of solution<\/strong>.<\/li>\n\n\n\n<li>Water is the solvent; hence, solvent mass = <strong>100 g &#8211; 70 g = 30 g = 0.030 kg<\/strong>.<\/li>\n\n\n\n<li>Molar masses: <strong>Isopropyl alcohol = 60.1 g\/mol, Ethyl alcohol = 46.1 g\/mol<\/strong>.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 2: Calculate Molality (m)<\/strong><\/h3>\n\n\n\n<p>[<br>m = \\frac{\\text{mass of solute}}{\\text{molar mass} \\times \\text{mass of solvent in kg}}<br>]<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>For Isopropyl Alcohol<\/strong>:<\/h4>\n\n\n\n<p>[<br>m = \\frac{70}{60.1 \\times 0.030} = 38.8 \\text{ mol\/kg}<br>]<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>For Ethyl Alcohol<\/strong>:<\/h4>\n\n\n\n<p>[<br>m = \\frac{70}{46.1 \\times 0.030} = 50.6 \\text{ mol\/kg}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 3: Calculate Freezing Point Depression<\/strong><\/h3>\n\n\n\n<p>[<br>\\Delta T_f = 1 \\times 1.86 \\times m<br>]<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>For Isopropyl Alcohol<\/strong>:<\/h4>\n\n\n\n<p>[<br>\\Delta T_f = 1.86 \\times 38.8 = 72.2^{\\circ}C<br>]<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>For Ethyl Alcohol<\/strong>:<\/h4>\n\n\n\n<p>[<br>\\Delta T_f = 1.86 \\times 50.6 = 94.1^{\\circ}C<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 4: Determine New Freezing Points<\/strong><\/h3>\n\n\n\n<p>Water\u2019s normal freezing point = <strong>0\u00b0C<\/strong>.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Isopropyl alcohol solution<\/strong>: ( 0 &#8211; 72.2 = -72.2^{\\circ}C )<\/li>\n\n\n\n<li><strong>Ethyl alcohol solution<\/strong>: ( 0 &#8211; 94.1 = -94.1^{\\circ}C )<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation (300 Words)<\/strong><\/h3>\n\n\n\n<p>Freezing point depression occurs when a solute is dissolved in a solvent, lowering its freezing temperature. This happens because the solute particles disrupt the solvent\u2019s ability to form a solid structure, requiring a lower temperature to freeze.<\/p>\n\n\n\n<p>For <strong>isopropyl alcohol (70%)<\/strong>, the large number of solute molecules in a small amount of water significantly decreases the freezing point to <strong>-72.2\u00b0C<\/strong>. Similarly, <strong>ethyl alcohol (70%)<\/strong> lowers the freezing point even more to <strong>-94.1\u00b0C<\/strong> due to its higher molality.<\/p>\n\n\n\n<p>This principle is widely used in <strong>antifreeze<\/strong> solutions, preventing liquids from freezing in cold temperatures. In real-life applications, <strong>alcohol-based solutions<\/strong> like rubbing alcohol remain in liquid form at sub-zero temperatures, making them useful for medical, industrial, and de-icing purposes.<\/p>\n\n\n\n<p>Now, I will generate an image illustrating <strong>freezing point depression<\/strong> for both alcohols.<\/p>\n\n\n\n<p>Here is the scientific illustration showing the freezing point depression of 70% isopropyl alcohol (-72.2\u00b0C) and 70% ethyl alcohol (-94.1\u00b0C) compared to pure water (0\u00b0C). The image includes thermometers, labeled test tubes, and a cold environment representation. Let me know if you need any modifications or further explanations!<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-1381.png\" alt=\"\" class=\"wp-image-194602\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Isopropyl Alcohol 70% Ethyl Alcohol 70% Calculate the freezing point depression The Correct Answer and Explanation is : To calculate the freezing point depression ((\\Delta T_f)) for 70% isopropyl alcohol and 70% ethyl alcohol, we use the freezing point depression equation: [\\Delta T_f = i \\cdot K_f \\cdot m] Where: Step 1: Define Given Data [&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-194601","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/194601","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=194601"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/194601\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=194601"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=194601"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=194601"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}