{"id":196102,"date":"2025-03-02T09:55:48","date_gmt":"2025-03-02T09:55:48","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=196102"},"modified":"2025-03-02T09:55:51","modified_gmt":"2025-03-02T09:55:51","slug":"acetone-is-the-main-ingredient-of-nail-polish-remover","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/02\/acetone-is-the-main-ingredient-of-nail-polish-remover\/","title":{"rendered":"Acetone is the main ingredient of nail polish remover"},"content":{"rendered":"\n
Acetone is the main ingredient of nail polish remover. A solution is made up by adding 35.0 ml acetone (density = 0.790 g\/ml)\u00c3\u201a\u00c2 to 50.0 ml of ethyl alcohol, \u00c3\u201a\u00c2 (density =0.789 g\/ml), Assuming volumes are additive. Calculate, Mass percent of acetone, the Volume percent of ethyl alcohol, and Mole fraction of acetone<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n
Let’s solve for the mass percent of acetone, the volume percent of ethyl alcohol, and the mole fraction of acetone in the solution.<\/p>\n\n\n\n
1. Mass Percent of Acetone<\/strong><\/h3>\n\n\n\n
The mass percent of acetone is calculated as: [ \\text{Mass percent of acetone} = \\frac{\\text{Mass of acetone}}{\\text{Total mass of solution}} \\times 100 ]<\/p>\n\n\n\n
Step 1: Calculate the mass of acetone<\/strong><\/p>\n\n\n\n
\n
Volume of acetone = 35.0 mL<\/li>\n\n\n\n
Density of acetone = 0.790 g\/mL [ \\text{Mass of acetone} = \\text{Volume} \\times \\text{Density} = 35.0 \\, \\text{mL} \\times 0.790 \\, \\text{g\/mL} = 27.65 \\, \\text{g} ]<\/li>\n<\/ul>\n\n\n\n
Step 2: Calculate the mass of ethyl alcohol<\/strong><\/p>\n\n\n\n
\n
Volume of ethyl alcohol = 50.0 mL<\/li>\n\n\n\n
Density of ethyl alcohol = 0.789 g\/mL [ \\text{Mass of ethyl alcohol} = \\text{Volume} \\times \\text{Density} = 50.0 \\, \\text{mL} \\times 0.789 \\, \\text{g\/mL} = 39.45 \\, \\text{g} ]<\/li>\n<\/ul>\n\n\n\n
Step 3: Calculate the total mass of the solution<\/strong> [ \\text{Total mass} = \\text{Mass of acetone} + \\text{Mass of ethyl alcohol} = 27.65 \\, \\text{g} + 39.45 \\, \\text{g} = 67.10 \\, \\text{g} ]<\/p>\n\n\n\n
Step 4: Calculate the mass percent of acetone<\/strong> [ \\text{Mass percent of acetone} = \\frac{27.65 \\, \\text{g}}{67.10 \\, \\text{g}} \\times 100 = 41.2\\% ]<\/p>\n\n\n\n
2. Volume Percent of Ethyl Alcohol<\/strong><\/h3>\n\n\n\n
The volume percent of ethyl alcohol is calculated as: [ \\text{Volume percent of ethyl alcohol} = \\frac{\\text{Volume of ethyl alcohol}}{\\text{Total volume of solution}} \\times 100 ]<\/p>\n\n\n\n
Step 1: Calculate the total volume of the solution<\/strong><\/p>\n\n\n\n
Step 2: Calculate the volume percent of ethyl alcohol<\/strong> [ \\text{Volume percent of ethyl alcohol} = \\frac{50.0 \\, \\text{mL}}{85.0 \\, \\text{mL}} \\times 100 = 58.8\\% ]<\/p>\n\n\n\n
3. Mole Fraction of Acetone<\/strong><\/h3>\n\n\n\n
The mole fraction of acetone is calculated as: [ \\text{Mole fraction of acetone} = \\frac{\\text{moles of acetone}}{\\text{moles of acetone} + \\text{moles of ethyl alcohol}} ]<\/p>\n\n\n\n
Step 1: Calculate the moles of acetone<\/strong><\/p>\n\n\n\n
\n
Molar mass of acetone (C\u2083H\u2086O) = 58.08 g\/mol [ \\text{moles of acetone} = \\frac{\\text{mass of acetone}}{\\text{molar mass of acetone}} = \\frac{27.65 \\, \\text{g}}{58.08 \\, \\text{g\/mol}} = 0.476 \\, \\text{mol} ]<\/li>\n<\/ul>\n\n\n\n
Step 2: Calculate the moles of ethyl alcohol<\/strong><\/p>\n\n\n\n
\n
Molar mass of ethyl alcohol (C\u2082H\u2085OH) = 46.07 g\/mol [ \\text{moles of ethyl alcohol} = \\frac{\\text{mass of ethyl alcohol}}{\\text{molar mass of ethyl alcohol}} = \\frac{39.45 \\, \\text{g}}{46.07 \\, \\text{g\/mol}} = 0.857 \\, \\text{mol} ]<\/li>\n<\/ul>\n\n\n\n
Step 3: Calculate the mole fraction of acetone<\/strong> [ \\text{Mole fraction of acetone} = \\frac{0.476 \\, \\text{mol}}{0.476 \\, \\text{mol} + 0.857 \\, \\text{mol}} = \\frac{0.476}{1.333} = 0.357 ]<\/p>\n\n\n\n
Final Answers:<\/h3>\n\n\n\n
\n
Mass percent of acetone<\/strong> = 41.2%<\/li>\n\n\n\n
Volume percent of ethyl alcohol<\/strong> = 58.8%<\/li>\n\n\n\n
Mole fraction of acetone<\/strong> = 0.357<\/li>\n<\/ul>\n\n\n\n
I’ll now generate an image that shows the steps and results in a clear and visual format for better understanding.<\/p>\n\n\n\n<\/figure>\n\n\n\n
Here is a visual representation of the steps for calculating mass percent, volume percent, and mole fraction. It includes the formulas, the calculations, and the final answers for each. I hope this helps in understanding the process more clearly! Let me know if you need further assistance.<\/p>\n","protected":false},"excerpt":{"rendered":"
Acetone is the main ingredient of nail polish remover. A solution is made up by adding 35.0 ml acetone (density = 0.790 g\/ml)\u00c3\u201a\u00c2 to 50.0 ml of ethyl alcohol, \u00c3\u201a\u00c2 (density =0.789 g\/ml), Assuming volumes are additive. Calculate, Mass percent of acetone, the Volume percent of ethyl alcohol, and Mole fraction of acetone The Correct […]<\/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-196102","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196102","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=196102"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196102\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=196102"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=196102"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=196102"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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