{"id":227307,"date":"2025-06-06T04:41:22","date_gmt":"2025-06-06T04:41:22","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=227307"},"modified":"2025-06-06T04:41:24","modified_gmt":"2025-06-06T04:41:24","slug":"12-0-g-nahso4-mm-120-1-g-mol-is-dissolved-in-water-to-make-a-2-00-l-solution","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/06\/12-0-g-nahso4-mm-120-1-g-mol-is-dissolved-in-water-to-make-a-2-00-l-solution\/","title":{"rendered":"12.0 g NaHSO4 (MM: 120.1 g\/mol) is dissolved in water to make a 2.00 L solution."},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">12.0 g NaHSO4 (MM: 120.1 g\/mol) is dissolved in water to make a 2.00 L solution. What is the molarity of the resulting NaHSO4 solution?<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>The Correct Answer and Explanation is:<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">To calculate the molarity of the NaHSO\u2084 solution, we use the <strong>molarity formula<\/strong>: Molarity&nbsp;(M)=moles&nbsp;of&nbsp;soluteliters&nbsp;of&nbsp;solution\\text{Molarity (M)} = \\frac{\\text{moles of solute}}{\\text{liters of solution}}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 1: Calculate moles of NaHSO\u2084<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of NaHSO\u2084 = <strong>12.0 g<\/strong><\/li>\n\n\n\n<li>Molar mass (MM) of NaHSO\u2084 = <strong>120.1 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Moles&nbsp;of&nbsp;NaHSO\u2084=12.0&nbsp;g120.1&nbsp;g\/mol\u22480.0999&nbsp;mol\\text{Moles of NaHSO\u2084} = \\frac{12.0\\ \\text{g}}{120.1\\ \\text{g\/mol}} \\approx 0.0999\\ \\text{mol}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 2: Use volume of solution<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume of solution = <strong>2.00 L<\/strong><\/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\"><strong>Step 3: Plug into the molarity formula<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Molarity=0.0999&nbsp;mol2.00&nbsp;L=0.04995&nbsp;mol\/L\\text{Molarity} = \\frac{0.0999\\ \\text{mol}}{2.00\\ \\text{L}} = 0.04995\\ \\text{mol\/L} Final&nbsp;Answer:0.0500&nbsp;M&nbsp;(rounded&nbsp;to&nbsp;3&nbsp;significant&nbsp;figures)\\text{Final Answer:} \\boxed{0.0500\\ \\text{M}} \\ (\\text{rounded to 3 significant figures})<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Molarity is a common way to express the concentration of a solution in chemistry. It tells us how many moles of solute are present in each liter of solution. In this case, sodium bisulfate (NaHSO\u2084) is the solute, and water is the solvent.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">First, to determine the number of moles of NaHSO\u2084, we use its molar mass, which is the mass of one mole of the compound. This is calculated by summing the atomic masses of its constituent elements: sodium (Na), hydrogen (H), sulfur (S), and oxygen (O). This gives a molar mass of 120.1 g\/mol. By dividing the given mass of NaHSO\u2084 (12.0 g) by its molar mass, we find the number of moles, which is approximately 0.0999 mol.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Next, since the solute is dissolved in enough water to make <strong>2.00 liters<\/strong> of solution, we divide the number of moles by this volume to find the molarity. This gives a value of approximately 0.04995 mol\/L. Following standard rules of significant figures (based on the 3 significant figures in the given values), we round the answer to <strong>0.0500 M<\/strong>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This result tells us that for every liter of this solution, there are 0.0500 moles of NaHSO\u2084 dissolved. Knowing the molarity is essential in chemical reactions and lab work, especially when dealing with acids and bases, where exact concentrations determine reaction rates and product yields.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner4-519.jpeg\" alt=\"\" class=\"wp-image-227308\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>12.0 g NaHSO4 (MM: 120.1 g\/mol) is dissolved in water to make a 2.00 L solution. What is the molarity of the resulting NaHSO4 solution? The Correct Answer and Explanation is: To calculate the molarity of the NaHSO\u2084 solution, we use the molarity formula: Molarity&nbsp;(M)=moles&nbsp;of&nbsp;soluteliters&nbsp;of&nbsp;solution\\text{Molarity (M)} = \\frac{\\text{moles of solute}}{\\text{liters of solution}} Step 1: Calculate [&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-227307","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/227307","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=227307"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/227307\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=227307"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=227307"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=227307"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}