{"id":183834,"date":"2025-01-17T06:40:48","date_gmt":"2025-01-17T06:40:48","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=183834"},"modified":"2025-01-17T06:40:50","modified_gmt":"2025-01-17T06:40:50","slug":"suppose-you-had-58-44-g-of-nacl-and-you-dissolved-it-in-exactly-2-00-l-of-water","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/17\/suppose-you-had-58-44-g-of-nacl-and-you-dissolved-it-in-exactly-2-00-l-of-water\/","title":{"rendered":"Suppose you had 58.44 g of NaCl and you dissolved it in exactly 2.00 L of water"},"content":{"rendered":"\n<p>Suppose you had 58.44 g of NaCl and you dissolved it in exactly 2.00 L of water. What would be the molality? 2. Calculate the molarity of 25.0 grams of KBr dissolved to 750.0 mL. 3. Eighty grams of glucose (C6H12O6, MW = 180. g\/mol) is dissolved in enough water to make 1.00 L of solution. What is its molarity? 4. When 2.00 grams of KMnO4 (MW = 158.0 g\/mol) is dissolved into 100.0 mL of solution, what normality results? How many grams of KMnO4 are needed to make 500.0 mL of a 0.200 N solution? 5. Ten grams of acetic acid (CH3COOH) is dissolved in 500.0 mL of solution. What is the molarity?  6. How many mL of solution will result when 15.0 g of H2SO4 is dissolved to make a 0.200 M solution? 7. Convert 0.169 Molar HC2H3O2 to % (m\/v). 8. One gram of sodium hydroxide pellets is dissolved in two hundred-fifty milliliters water. What is the molarity, molality and normality of the solution? Assume that the density of solution to be approximately that of the water and the density of sodium hydroxide pellets to be 2.13 g\/cm\u00b3.<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Solutions and Explanations:<\/h3>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Molality of NaCl solution:<\/strong> <strong>Given:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of NaCl = 58.44 g<\/li>\n\n\n\n<li>Volume of water = 2.00 L \u2248 2000 g (assume density of water = 1 g\/mL) <strong>Steps:<\/strong><\/li>\n\n\n\n<li>Moles of NaCl = ( \\frac{58.44 \\, \\text{g}}{58.44 \\, \\text{g\/mol}} = 1.00 \\, \\text{mol} )<\/li>\n\n\n\n<li>Molality ((m)) = ( \\frac{\\text{moles of solute}}{\\text{mass of solvent in kg}} )<\/li>\n\n\n\n<li>( m = \\frac{1.00}{2.00} = 0.500 \\, \\text{mol\/kg} ) <strong>Answer:<\/strong> ( 0.500 \\, m )<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<ol start=\"2\" class=\"wp-block-list\">\n<li><strong>Molarity of KBr solution:<\/strong> <strong>Given:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of KBr = 25.0 g<\/li>\n\n\n\n<li>Volume of solution = 750.0 mL = 0.750 L<\/li>\n\n\n\n<li>Molar mass of KBr = 119.0 g\/mol <strong>Steps:<\/strong><\/li>\n\n\n\n<li>Moles of KBr = ( \\frac{25.0}{119.0} = 0.210 \\, \\text{mol} )<\/li>\n\n\n\n<li>Molarity ((M)) = ( \\frac{\\text{moles of solute}}{\\text{volume of solution in L}} )<\/li>\n\n\n\n<li>( M = \\frac{0.210}{0.750} = 0.280 \\, \\text{M} ) <strong>Answer:<\/strong> ( 0.280 \\, M )<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<ol start=\"3\" class=\"wp-block-list\">\n<li><strong>Molarity of glucose solution:<\/strong> <strong>Given:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of glucose = 80.0 g<\/li>\n\n\n\n<li>Volume of solution = 1.00 L<\/li>\n\n\n\n<li>Molar mass of glucose = 180. g\/mol <strong>Steps:<\/strong><\/li>\n\n\n\n<li>Moles of glucose = ( \\frac{80.0}{180.} = 0.444 \\, \\text{mol} )<\/li>\n\n\n\n<li>Molarity ((M)) = ( \\frac{0.444}{1.00} = 0.444 \\, \\text{M} ) <strong>Answer:<\/strong> ( 0.444 \\, M )<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<ol start=\"4\" class=\"wp-block-list\">\n<li><strong>Normality of KMnO(_4) solution:<\/strong> <strong>Given:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of KMnO(_4) = 2.00 g<\/li>\n\n\n\n<li>Volume = 100.0 mL = 0.100 L<\/li>\n\n\n\n<li>Molar mass = 158.0 g\/mol<\/li>\n\n\n\n<li>Equivalent factor = 5 (KMnO(_4) in acidic medium) <strong>Steps:<\/strong><\/li>\n\n\n\n<li>Moles of KMnO(_4) = ( \\frac{2.00}{158.0} = 0.0127 \\, \\text{mol} )<\/li>\n\n\n\n<li>Normality ((N)) = ( \\text{Molarity} \\times \\text{Equivalent factor} )<\/li>\n\n\n\n<li>( N = \\frac{0.0127}{0.100} \\times 5 = 0.635 \\, N ) <strong>Grams needed for 0.200 N solution:<\/strong><\/li>\n\n\n\n<li>( 0.200 = \\frac{\\text{grams of KMnO}_4 \\times 5}{158.0 \\times 0.500} )<\/li>\n\n\n\n<li>Grams = ( 3.16 \\, \\text{g} ) <strong>Answer:<\/strong> ( N = 0.635, \\, \\text{Grams needed} = 3.16 \\, \\text{g} )<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<ol start=\"5\" class=\"wp-block-list\">\n<li><strong>Molarity of acetic acid:<\/strong> <strong>Given:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of CH(_3)COOH = 10.0 g<\/li>\n\n\n\n<li>Volume = 500.0 mL = 0.500 L<\/li>\n\n\n\n<li>Molar mass = 60.0 g\/mol <strong>Steps:<\/strong><\/li>\n\n\n\n<li>Moles = ( \\frac{10.0}{60.0} = 0.167 \\, \\text{mol} )<\/li>\n\n\n\n<li>( M = \\frac{0.167}{0.500} = 0.334 \\, \\text{M} ) <strong>Answer:<\/strong> ( 0.334 \\, M )<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<ol start=\"6\" class=\"wp-block-list\">\n<li><strong>Volume for H(_2)SO(_4) solution:<\/strong> <strong>Given:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of H(_2)SO(_4) = 15.0 g<\/li>\n\n\n\n<li>Molar mass = 98.0 g\/mol<\/li>\n\n\n\n<li>( M = 0.200 \\, \\text{M} ) <strong>Steps:<\/strong><\/li>\n\n\n\n<li>Moles = ( \\frac{15.0}{98.0} = 0.153 \\, \\text{mol} )<\/li>\n\n\n\n<li>Volume = ( \\frac{\\text{moles}}{M} = \\frac{0.153}{0.200} = 0.765 \\, \\text{L} ) <strong>Answer:<\/strong> ( 765 \\, \\text{mL} )<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<ol start=\"7\" class=\"wp-block-list\">\n<li><strong>Convert 0.169 M HC(_2)H(_3)O(_2) to % (m\/v):<\/strong> <strong>Given:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( M = 0.169 \\, \\text{M} )<\/li>\n\n\n\n<li>Molar mass = 60.0 g\/mol <strong>Steps:<\/strong><\/li>\n\n\n\n<li>( 0.169 \\, \\text{mol\/L} \\times 60.0 \\, \\text{g\/mol} = 10.14 \\, \\text{g\/L} )<\/li>\n\n\n\n<li>% (m\/v) = ( \\frac{10.14 \\, \\text{g}}{1000 \\, \\text{mL}} \\times 100 = 1.01\\% ) <strong>Answer:<\/strong> ( 1.01\\% \\, \\text{(m\/v)} )<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<ol start=\"8\" class=\"wp-block-list\">\n<li><strong>Sodium hydroxide solution:<\/strong> <strong>Given:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of NaOH = 1.00 g<\/li>\n\n\n\n<li>Volume = 250.0 mL = 0.250 L<\/li>\n\n\n\n<li>Molar mass = 40.0 g\/mol <strong>Steps:<\/strong><\/li>\n\n\n\n<li>Moles = ( \\frac{1.00}{40.0} = 0.025 \\, \\text{mol} )<\/li>\n\n\n\n<li>Molarity = ( \\frac{0.025}{0.250} = 0.100 \\, \\text{M} )<\/li>\n\n\n\n<li>Assume ( \\text{mass of solution} = 250 \\, \\text{g} ) (water density)<\/li>\n\n\n\n<li>Molality = ( \\frac{0.025}{0.250} = 0.100 \\, \\text{mol\/kg} )<\/li>\n\n\n\n<li>Normality ((N)) = Molarity (1 equivalent) <strong>Answer:<\/strong> ( 0.100 \\, M, \\, 0.100 \\, m, \\, 0.100 \\, N )<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The calculations follow the basic definitions of molarity, molality, and normality, with the molar mass and volume conversions ensuring accurate results.<\/li>\n\n\n\n<li>Assumptions such as water density simplify calculations without significantly affecting accuracy for dilute solutions.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Suppose you had 58.44 g of NaCl and you dissolved it in exactly 2.00 L of water. What would be the molality? 2. Calculate the molarity of 25.0 grams of KBr dissolved to 750.0 mL. 3. Eighty grams of glucose (C6H12O6, MW = 180. g\/mol) is dissolved in enough water to make 1.00 L of [&hellip;]<\/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-183834","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183834","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=183834"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183834\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=183834"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=183834"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=183834"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}