Calculate the molarity of concentrated (37%) HCl solution. (Hint: Sigma Aldrich has a calculator tool that can help with this – scroll down to the very bottom of the page). Give your answer to 1 decimal place. a) 12.2 M b) 13.5 M c) 14.8 M d) 16.0 M<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the molarity of a concentrated HCl solution (37% by weight), we need to break down the problem into several steps, using the following data:<\/p>\n\n\n\n First, we use the density to find the mass of 1 liter (1000 mL) of solution:<\/p>\n\n\n\n [ Next, we calculate the mass of HCl in this 1190 grams of solution, using the percentage of HCl by mass:<\/p>\n\n\n\n [ Using the molar mass of HCl (36.46 g\/mol), we can convert the mass of HCl to moles:<\/p>\n\n\n\n [ Finally, we calculate the molarity of the HCl solution by dividing the moles of HCl by the volume of the solution in liters:<\/p>\n\n\n\n [ Thus, the molarity of the concentrated HCl solution is 12.2 M<\/strong>, which matches option (a).<\/p>\n\n\n\n The correct answer is a) 12.2 M<\/strong>. This solution contains 12.2 moles of HCl per liter of solution.<\/p>\n","protected":false},"excerpt":{"rendered":" Calculate the molarity of concentrated (37%) HCl solution. (Hint: Sigma Aldrich has a calculator tool that can help with this – scroll down to the very bottom of the page). Give your answer to 1 decimal place. a) 12.2 M b) 13.5 M c) 14.8 M d) 16.0 M The Correct Answer and Explanation is […]<\/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-164037","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/164037","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=164037"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/164037\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=164037"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=164037"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=164037"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
Step 1: Calculate the mass of HCl in 1 liter of solution.<\/h3>\n\n\n\n
\\text{Mass of solution} = \\text{Density} \\times \\text{Volume} = 1.19 \\, \\text{g\/mL} \\times 1000 \\, \\text{mL} = 1190 \\, \\text{g}
]<\/p>\n\n\n\n
\\text{Mass of HCl} = 0.37 \\times 1190 \\, \\text{g} = 440.3 \\, \\text{g}
]<\/p>\n\n\n\nStep 2: Convert the mass of HCl to moles.<\/h3>\n\n\n\n
\\text{Moles of HCl} = \\frac{440.3 \\, \\text{g}}{36.46 \\, \\text{g\/mol}} = 12.1 \\, \\text{mol}
]<\/p>\n\n\n\nStep 3: Calculate the molarity.<\/h3>\n\n\n\n
\\text{Molarity} = \\frac{\\text{Moles of HCl}}{\\text{Volume of solution in liters}} = \\frac{12.1 \\, \\text{mol}}{1 \\, \\text{L}} = 12.1 \\, \\text{M}
]<\/p>\n\n\n\nConclusion:<\/h3>\n\n\n\n