{"id":35338,"date":"2025-06-24T09:48:48","date_gmt":"2025-06-24T09:48:48","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=35338"},"modified":"2025-06-24T09:48:49","modified_gmt":"2025-06-24T09:48:49","slug":"the-heat-of-vaporization-of-rubbing-alcohol-2-propanol-is-approximately-42-kj-mol","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/the-heat-of-vaporization-of-rubbing-alcohol-2-propanol-is-approximately-42-kj-mol\/","title":{"rendered":"The heat of vaporization of rubbing alcohol (2-propanol) is approximately 42 kJ\/mol."},"content":{"rendered":"\n<p>The heat of vaporization of rubbing alcohol (2-propanol) is approximately 42 kJ\/mol. How much heat is required to evaporate 1.5 g of rubbing alcohol?<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To determine the amount of heat required to evaporate 1.5 grams of rubbing alcohol (2-propanol), we follow these steps:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Determine the molar mass of 2-propanol<\/h3>\n\n\n\n<p>The chemical formula for 2-propanol is <strong>C\u2083H\u2088O<\/strong>.<br>Molar mass is calculated as:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon (C): 12.01 \u00d7 3 = 36.03 g\/mol<\/li>\n\n\n\n<li>Hydrogen (H): 1.008 \u00d7 8 = 8.064 g\/mol<\/li>\n\n\n\n<li>Oxygen (O): 16.00 \u00d7 1 = 16.00 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>Total molar mass = 36.03 + 8.064 + 16.00 = <strong>60.094 g\/mol<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Convert mass to moles<\/h3>\n\n\n\n<p>Moles = mass \u00f7 molar mass = 1.5 g \u00f7 60.094 g\/mol \u2248 <strong>0.02496 mol<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Use heat of vaporization to find heat required<\/h3>\n\n\n\n<p>Heat = moles \u00d7 heat of vaporization<br>Heat = 0.02496 mol \u00d7 42 kJ\/mol \u2248 <strong>1.048 kJ<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p><strong>Approximately 1.05 kJ<\/strong> of heat is required to evaporate 1.5 grams of rubbing alcohol.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation :<\/h3>\n\n\n\n<p>Understanding how much heat is needed to evaporate a given amount of a substance involves a key concept in thermodynamics known as the heat of vaporization. The heat of vaporization is the amount of energy required to convert one mole of a liquid into vapor at a constant temperature and pressure, without changing its temperature. For 2-propanol, also known as rubbing alcohol, this value is approximately 42 kilojoules per mole.<\/p>\n\n\n\n<p>To calculate the total heat required, the first step is converting the given mass of 2-propanol to moles using its molar mass. The molar mass is determined by adding up the atomic masses of all atoms in a molecule of 2-propanol. Once the number of moles is known, we multiply it by the heat of vaporization. This gives the total energy required to vaporize that amount of alcohol.<\/p>\n\n\n\n<p>In this case, 1.5 grams of 2-propanol corresponds to approximately 0.025 moles. When this is multiplied by the heat of vaporization, we get the total heat required for evaporation, which is about 1.05 kilojoules. This value represents the energy that must be absorbed by the liquid alcohol to overcome intermolecular forces and change into the vapor phase.<\/p>\n\n\n\n<p>This type of calculation is useful in chemistry and engineering, particularly when designing processes that involve phase changes like distillation or evaporation. It helps determine energy requirements and efficiency of thermal systems.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-669.jpeg\" alt=\"\" class=\"wp-image-35339\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-669.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-669-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-669-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The heat of vaporization of rubbing alcohol (2-propanol) is approximately 42 kJ\/mol. How much heat is required to evaporate 1.5 g of rubbing alcohol? The Correct Answer and Explanation is: To determine the amount of heat required to evaporate 1.5 grams of rubbing alcohol (2-propanol), we follow these steps: Step 1: Determine the molar mass [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-35338","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35338","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=35338"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35338\/revisions"}],"predecessor-version":[{"id":35340,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35338\/revisions\/35340"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=35338"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=35338"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=35338"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}