Calculate the molarity of (mol\/L) of 3% hydrogen peroxide and record it in Data Table 2. The term \u201c3% concentration\u201d is a weight-to-weight measurement, which means that there are 3 g solute per 100 g solvent. The molar mass of hydrogen peroxide is 34.01 g\/mol and the density of the solution is 1.01 g\/mL.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n The molarity of a 3% hydrogen peroxide solution is approximately 0.891 mol\/L<\/strong>.<\/p>\n\n\n\n To calculate the molarity (mol\/L) of a 3% hydrogen peroxide solution, we follow these steps:<\/p>\n\n\n\n This gives us the molarity of the hydrogen peroxide solution as approximately 0.891 mol\/L<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":" Calculate the molarity of (mol\/L) of 3% hydrogen peroxide and record it in Data Table 2. The term \u201c3% concentration\u201d is a weight-to-weight measurement, which means that there are 3 g solute per 100 g solvent. The molar mass of hydrogen peroxide is 34.01 g\/mol and the density of the solution is 1.01 g\/mL. The […]<\/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-196062","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196062","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=196062"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196062\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=196062"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=196062"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=196062"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Explanation:<\/h3>\n\n\n\n
\n
The 3% concentration means that there are 3 grams of hydrogen peroxide (H\u2082O\u2082) per 100 grams of solution. For simplicity, we assume a 1-liter solution (1000 mL). Given that the density of the solution is 1.01 g\/mL, the mass of the entire solution in 1 liter is calculated as: [
\\text{Mass of solution} = \\text{density} \\times \\text{volume} = 1.01 \\, \\text{g\/mL} \\times 1000 \\, \\text{mL} = 1010 \\, \\text{grams}
]<\/li>\n\n\n\n
Since the solution is 3% H\u2082O\u2082 by weight, the mass of hydrogen peroxide in the 1-liter solution is: [
\\text{Mass of H\u2082O\u2082} = \\frac{3}{100} \\times 1010 \\, \\text{g} = 30.3 \\, \\text{g}
]<\/li>\n\n\n\n
To calculate the number of moles of H\u2082O\u2082, we use its molar mass, which is 34.01 g\/mol. Thus, the number of moles of H\u2082O\u2082 in the solution is: [
\\text{Moles of H\u2082O\u2082} = \\frac{30.3 \\, \\text{g}}{34.01 \\, \\text{g\/mol}} \\approx 0.891 \\, \\text{mol}
]<\/li>\n\n\n\n
Molarity is defined as moles of solute per liter of solution. Therefore, for a 1-liter solution: [
\\text{Molarity} = \\frac{0.891 \\, \\text{mol}}{1 \\, \\text{L}} = 0.891 \\, \\text{mol\/L}
]<\/li>\n<\/ol>\n\n\n\n