{"id":211788,"date":"2025-05-05T07:37:14","date_gmt":"2025-05-05T07:37:14","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=211788"},"modified":"2025-05-05T07:37:16","modified_gmt":"2025-05-05T07:37:16","slug":"the-density-of-dichloromethane-an-organic-solvent-is-1-33-g-ml-at-4c","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/05\/the-density-of-dichloromethane-an-organic-solvent-is-1-33-g-ml-at-4c\/","title":{"rendered":"The density of dichloromethane, an organic solvent, is 1.33 g\/mL at 4\u00b0C"},"content":{"rendered":"\n<p>The density of dichloromethane, an organic solvent, is 1.33 g\/mL at 4\u00b0C. How many dichloromethane molecules are present in 2.35 mL of dichloromethane at this temperature? (Molar Mass of dichloromethane: 84.93 g\/mol)<\/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 determine how many molecules of dichloromethane are present in 2.35 mL of the liquid at 4\u00b0C, we need to follow these steps:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Convert volume of dichloromethane to mass<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The density of dichloromethane = 1.33 g\/mL<\/li>\n\n\n\n<li>Volume of dichloromethane = 2.35 mL<\/li>\n<\/ul>\n\n\n\n<p>We can calculate the mass (m) of dichloromethane using the formula:<\/p>\n\n\n\n<p>$$<br>\\text{Mass} = \\text{Density} \\times \\text{Volume}<br>$$<\/p>\n\n\n\n<p>Substituting the given values:<\/p>\n\n\n\n<p>$$<br>\\text{Mass} = 1.33 \\, \\text{g\/mL} \\times 2.35 \\, \\text{mL} = 3.1305 \\, \\text{g}<br>$$<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Convert the mass of dichloromethane to moles<\/h3>\n\n\n\n<p>Now that we have the mass of dichloromethane, we can convert it to moles using the molar mass. The molar mass of dichloromethane is given as 84.93 g\/mol.<\/p>\n\n\n\n<p>To find the number of moles (n), we use the equation:<\/p>\n\n\n\n<p>$$<br>\\text{Moles} = \\frac{\\text{Mass}}{\\text{Molar Mass}}<br>$$<\/p>\n\n\n\n<p>Substituting the values:<\/p>\n\n\n\n<p>$$<br>\\text{Moles} = \\frac{3.1305 \\, \\text{g}}{84.93 \\, \\text{g\/mol}} = 0.0369 \\, \\text{mol}<br>$$<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Convert moles to molecules<\/h3>\n\n\n\n<p>To find the number of molecules, we use Avogadro&#8217;s number, which is $6.022 \\times 10^{23}$ molecules per mole.<\/p>\n\n\n\n<p>$$<br>\\text{Number of molecules} = \\text{Moles} \\times \\text{Avogadro&#8217;s number}<br>$$<\/p>\n\n\n\n<p>Substituting the values:<\/p>\n\n\n\n<p>$$<br>\\text{Number of molecules} = 0.0369 \\, \\text{mol} \\times 6.022 \\times 10^{23} \\, \\text{molecules\/mol}<br>$$<\/p>\n\n\n\n<p>$$<br>\\text{Number of molecules} = 2.22 \\times 10^{22} \\, \\text{molecules}<br>$$<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>The number of dichloromethane molecules present in 2.35 mL of dichloromethane at 4\u00b0C is approximately $2.22 \\times 10^{22}$ molecules.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Density to Mass Conversion:<\/strong> The density tells us how much mass is present in a given volume of the liquid. By multiplying the density (1.33 g\/mL) by the volume (2.35 mL), we get the mass of the liquid.<\/li>\n\n\n\n<li><strong>Mass to Moles Conversion:<\/strong> Using the molar mass of dichloromethane (84.93 g\/mol), we can convert the mass into moles. This tells us how many moles of dichloromethane are present in the given volume.<\/li>\n\n\n\n<li><strong>Moles to Molecules Conversion:<\/strong> Finally, by multiplying the number of moles by Avogadro\u2019s number, we can calculate the number of individual molecules in the sample.<\/li>\n<\/ol>\n\n\n\n<p>These steps allow us to move from a macroscopic measurement (volume) to a microscopic understanding (molecular count), illustrating the relationship between mass, moles, and molecules.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The density of dichloromethane, an organic solvent, is 1.33 g\/mL at 4\u00b0C. How many dichloromethane molecules are present in 2.35 mL of dichloromethane at this temperature? (Molar Mass of dichloromethane: 84.93 g\/mol) The correct answer and explanation is : To determine how many molecules of dichloromethane are present in 2.35 mL of the liquid at [&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-211788","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/211788","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=211788"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/211788\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=211788"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=211788"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=211788"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}