A Solution Of Hydrogen Peroxide Is 15.2% By Mass. What Is The Molarity Of The Solution? Assume That The Solution Has A Density Of 1.01g\/Ml. Density Of 1.01g\/ML.<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To find the molarity of the hydrogen peroxide (H\u2082O\u2082) solution, we need to follow these steps:<\/p>\n\n\n\n The mass percentage indicates that in 100 g of solution, there are 15.2 g of H\u2082O\u2082. Therefore, the amount of hydrogen peroxide in 100 g of solution is:<\/p>\n\n\n\n [ The remaining mass of the solution (100 g – 15.2 g) is water, which is 84.8 g.<\/p>\n\n\n\n The density of the solution is given as 1.01 g\/mL. Density is defined as:<\/p>\n\n\n\n [ Rearranging to find the volume:<\/p>\n\n\n\n [ Next, we calculate the moles of H\u2082O\u2082 in the solution. Using the molar mass of H\u2082O\u2082 (34.01 g\/mol), the moles of H\u2082O\u2082 in 15.2 g of the solution is:<\/p>\n\n\n\n [ Finally, molarity (M) is defined as the number of moles of solute per liter of solution. Using the volume of the solution from Step 2:<\/p>\n\n\n\n [ The molarity of the hydrogen peroxide solution is approximately 4.51 M<\/strong>.<\/p>\n\n\n\n This process involves several steps to ensure accurate conversion of mass to moles, then using the solution’s volume to determine molarity. We used the density to find the volume of the solution and used the molar mass to convert the mass of hydrogen peroxide into moles.<\/p>\n","protected":false},"excerpt":{"rendered":" A Solution Of Hydrogen Peroxide Is 15.2% By Mass. What Is The Molarity Of The Solution? Assume That The Solution Has A Density Of 1.01g\/Ml. Density Of 1.01g\/ML. The correct answer and explanation is : To find the molarity of the hydrogen peroxide (H\u2082O\u2082) solution, we need to follow these steps: Given: Step 1: Understanding […]<\/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-209989","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/209989","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=209989"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/209989\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=209989"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=209989"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=209989"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Given:<\/h3>\n\n\n\n
\n
Step 1: Understanding the mass percentage<\/h3>\n\n\n\n
\\text{Mass of H\u2082O\u2082} = 15.2 \\, \\text{g}
]<\/p>\n\n\n\nStep 2: Find the volume of the solution<\/h3>\n\n\n\n
\\text{Density} = \\frac{\\text{Mass of solution}}{\\text{Volume of solution}}
]<\/p>\n\n\n\n
\\text{Volume of solution} = \\frac{\\text{Mass of solution}}{\\text{Density}} = \\frac{100 \\, \\text{g}}{1.01 \\, \\text{g\/mL}} \\approx 99.01 \\, \\text{mL} = 0.09901 \\, \\text{L}
]<\/p>\n\n\n\nStep 3: Calculate moles of H\u2082O\u2082<\/h3>\n\n\n\n
\\text{Moles of H\u2082O\u2082} = \\frac{\\text{Mass of H\u2082O\u2082}}{\\text{Molar mass of H\u2082O\u2082}} = \\frac{15.2 \\, \\text{g}}{34.01 \\, \\text{g\/mol}} \\approx 0.447 \\, \\text{mol}
]<\/p>\n\n\n\nStep 4: Calculate the molarity<\/h3>\n\n\n\n
\\text{Molarity} = \\frac{\\text{Moles of H\u2082O\u2082}}{\\text{Volume of solution in liters}} = \\frac{0.447 \\, \\text{mol}}{0.09901 \\, \\text{L}} \\approx 4.51 \\, \\text{M}
]<\/p>\n\n\n\nConclusion:<\/h3>\n\n\n\n