Calculate the molarity of a 30.0% hydrogen peroxide solution.The reported density of 30% hydrogen peroxide is 1.135 g\/cm3.(H2O2= 34.0147 g\/mol) a) How many grams of solution in 1L? b) howmany grams in hydrogen peroxide are in 1 L? c) What is the molarityand molality of the solution?<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the molarity and molality of a 30.0% hydrogen peroxide solution, we\u2019ll go through a series of steps using the given data:<\/p>\n\n\n\n Let\u2019s calculate each part step by step.<\/p>\n\n\n\n We\u2019re given the density of the solution as 1.135 g\/cm\u00b3. Since 1 L = 1000 cm\u00b3, we can calculate the mass of 1 L of solution: Since the solution is 30.0% by mass hydrogen peroxide, the mass of H\u2082O\u2082 in 1 L of solution is: Molarity is defined as moles of solute per liter of solution. To calculate the molarity, we first need to convert the mass of hydrogen peroxide to moles: Molality is defined as moles of solute per kilogram of solvent. We need to find the mass of the solvent, which is the mass of water in the solution. The mass of water is: Calculate the molarity of a 30.0% hydrogen peroxide solution.The reported density of 30% hydrogen peroxide is 1.135 g\/cm3.(H2O2= 34.0147 g\/mol) a) How many grams of solution in 1L? b) howmany grams in hydrogen peroxide are in 1 L? c) What is the molarityand molality of the solution? The Correct Answer and Explanation is : To […]<\/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-196064","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196064","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=196064"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196064\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=196064"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=196064"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=196064"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
a) How many grams of solution in 1 L?<\/h3>\n\n\n\n
[
\\text{Mass of 1 L of solution} = \\text{Density} \\times \\text{Volume}
]
[
\\text{Mass of 1 L of solution} = 1.135 \\, \\text{g\/cm}^3 \\times 1000 \\, \\text{cm}^3 = 1135 \\, \\text{g}
]
Thus, 1 L of solution weighs 1135 grams.<\/p>\n\n\n\nb) How many grams of hydrogen peroxide are in 1 L?<\/h3>\n\n\n\n
[
\\text{Mass of H\u2082O\u2082} = \\frac{30.0}{100} \\times 1135 \\, \\text{g} = 340.5 \\, \\text{g}
]
So, there are 340.5 grams of hydrogen peroxide in 1 L of solution.<\/p>\n\n\n\nc) Molarity and Molality of the solution<\/h3>\n\n\n\n
Molarity:<\/h4>\n\n\n\n
[
\\text{Moles of H\u2082O\u2082} = \\frac{\\text{Mass of H\u2082O\u2082}}{\\text{Molar mass of H\u2082O\u2082}}
]
[
\\text{Moles of H\u2082O\u2082} = \\frac{340.5 \\, \\text{g}}{34.0147 \\, \\text{g\/mol}} = 10.0 \\, \\text{mol}
]
Since the solution is 1 L, the molarity (M) is:
[
\\text{Molarity} = \\frac{10.0 \\, \\text{mol}}{1 \\, \\text{L}} = 10.0 \\, \\text{M}
]<\/p>\n\n\n\nMolality:<\/h4>\n\n\n\n
[
\\text{Mass of water} = \\text{Mass of solution} – \\text{Mass of H\u2082O\u2082} = 1135 \\, \\text{g} – 340.5 \\, \\text{g} = 794.5 \\, \\text{g} = 0.7945 \\, \\text{kg}
]
Now, we can calculate the molality (m):
[
\\text{Molality} = \\frac{10.0 \\, \\text{mol}}{0.7945 \\, \\text{kg}} = 12.6 \\, \\text{mol\/kg}
]<\/p>\n\n\n\nSummary:<\/h3>\n\n\n\n
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