{"id":188550,"date":"2025-02-07T05:21:03","date_gmt":"2025-02-07T05:21:03","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=188550"},"modified":"2025-02-07T05:21:05","modified_gmt":"2025-02-07T05:21:05","slug":"ethylene-glycol-an-antifreeze-has-a-density-of-1-11-g-ml-at-20-c","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/07\/ethylene-glycol-an-antifreeze-has-a-density-of-1-11-g-ml-at-20-c\/","title":{"rendered":"Ethylene glycol, an antifreeze. has a density of 1.11 g\/mL at 20 \u00b0C"},"content":{"rendered":"\n<p>Ethylene glycol, an antifreeze. has a density of 1.11 g\/mL at 20 \u00b0C. <\/p>\n\n\n\n<p>a) What is the mass, in grams of 452 mL ethylene glycol? <\/p>\n\n\n\n<p>b) What is the mass in kilograms of 18.6 L ethylene glycol? <\/p>\n\n\n\n<p>c) What is the volume, in milliliters, occupied by 65.0 g ethylene glycol?<\/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 solve these problems, we can use the basic density equation:<\/p>\n\n\n\n<p>[<br>\\text{Density} = \\frac{\\text{Mass}}{\\text{Volume}}<br>]<\/p>\n\n\n\n<p>This can be rearranged to solve for mass, volume, or density depending on the given information:<\/p>\n\n\n\n<p>[<br>\\text{Mass} = \\text{Density} \\times \\text{Volume}<br>]<br>[<br>\\text{Volume} = \\frac{\\text{Mass}}{\\text{Density}}<br>]<\/p>\n\n\n\n<p>Given the density of ethylene glycol as 1.11 g\/mL at 20 \u00b0C, we can now proceed with each part of the question.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">a) What is the mass in grams of 452 mL of ethylene glycol?<\/h3>\n\n\n\n<p>We are given the volume (452 mL) and the density (1.11 g\/mL), so we can use the formula for mass:<\/p>\n\n\n\n<p>[<br>\\text{Mass} = 1.11 \\, \\text{g\/mL} \\times 452 \\, \\text{mL} = 501.72 \\, \\text{g}<br>]<\/p>\n\n\n\n<p>So, the mass of 452 mL of ethylene glycol is <strong>501.72 grams<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">b) What is the mass in kilograms of 18.6 L of ethylene glycol?<\/h3>\n\n\n\n<p>First, we need to convert the volume from liters to milliliters (1 L = 1000 mL). So:<\/p>\n\n\n\n<p>[<br>18.6 \\, \\text{L} = 18.6 \\times 1000 = 18,600 \\, \\text{mL}<br>]<\/p>\n\n\n\n<p>Now we can calculate the mass using the density formula:<\/p>\n\n\n\n<p>[<br>\\text{Mass} = 1.11 \\, \\text{g\/mL} \\times 18,600 \\, \\text{mL} = 20,646 \\, \\text{g}<br>]<\/p>\n\n\n\n<p>To convert grams to kilograms (1 kg = 1000 g):<\/p>\n\n\n\n<p>[<br>\\text{Mass} = \\frac{20,646 \\, \\text{g}}{1000} = 20.646 \\, \\text{kg}<br>]<\/p>\n\n\n\n<p>So, the mass of 18.6 L of ethylene glycol is <strong>20.646 kilograms<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">c) What is the volume in milliliters occupied by 65.0 g of ethylene glycol?<\/h3>\n\n\n\n<p>We are given the mass (65.0 g) and the density (1.11 g\/mL), and we need to solve for the volume. Using the formula for volume:<\/p>\n\n\n\n<p>[<br>\\text{Volume} = \\frac{\\text{Mass}}{\\text{Density}} = \\frac{65.0 \\, \\text{g}}{1.11 \\, \\text{g\/mL}} = 58.65 \\, \\text{mL}<br>]<\/p>\n\n\n\n<p>So, the volume occupied by 65.0 g of ethylene glycol is <strong>58.65 milliliters<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Summary:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>a) 452 mL ethylene glycol has a mass of <strong>501.72 grams<\/strong>.<\/li>\n\n\n\n<li>b) 18.6 L ethylene glycol has a mass of <strong>20.646 kilograms<\/strong>.<\/li>\n\n\n\n<li>c) 65.0 g of ethylene glycol occupies <strong>58.65 milliliters<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p>These calculations are based on applying the density formula, using unit conversions when necessary, and ensuring consistent units throughout the problem.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Ethylene glycol, an antifreeze. has a density of 1.11 g\/mL at 20 \u00b0C. a) What is the mass, in grams of 452 mL ethylene glycol? b) What is the mass in kilograms of 18.6 L ethylene glycol? c) What is the volume, in milliliters, occupied by 65.0 g ethylene glycol? The Correct Answer and Explanation [&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-188550","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188550","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=188550"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188550\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=188550"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=188550"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=188550"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}