{"id":258690,"date":"2025-07-18T08:53:49","date_gmt":"2025-07-18T08:53:49","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=258690"},"modified":"2025-07-18T08:53:51","modified_gmt":"2025-07-18T08:53:51","slug":"0-274-g-manganese-chlorate-mnclo32-is-dissolved-in-water-to-make-100-0-ml-of-solution","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/18\/0-274-g-manganese-chlorate-mnclo32-is-dissolved-in-water-to-make-100-0-ml-of-solution\/","title":{"rendered":"0.274 g manganese chlorate (Mn(ClO3)2) is dissolved in water to make 100.0 mL of solution."},"content":{"rendered":"\n<p>0.274 g manganese chlorate (Mn(ClO3)2) is dissolved in water to make 100.0 mL of solution. Calculate the molarity of manganese chlorate in solution. (b) Calculate the mass percent of manganese chlorate in a solution made by dissolving 0.274 g Mn(ClO3)2 in 10.00 g water.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">(a) Molarity of manganese chlorate (Mn(ClO\u2083)\u2082)<\/h3>\n\n\n\n<p><strong>Step 1: Calculate the molar mass of Mn(ClO\u2083)\u2082.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The molar mass of Mn (Manganese) = 54.94 g\/mol<\/li>\n\n\n\n<li>The molar mass of Cl (Chlorine) = 35.45 g\/mol<\/li>\n\n\n\n<li>The molar mass of O (Oxygen) = 16.00 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>For Mn(ClO\u2083)\u2082:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The molar mass of Mn(ClO\u2083)\u2082 = 54.94 g\/mol + 2 \u00d7 (35.45 g\/mol + 3 \u00d7 16.00 g\/mol)<br>= 54.94 g\/mol + 2 \u00d7 (35.45 g\/mol + 48.00 g\/mol)<br>= 54.94 g\/mol + 2 \u00d7 83.45 g\/mol<br>= 54.94 g\/mol + 166.90 g\/mol<br>= 221.84 g\/mol<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 2: Calculate the moles of Mn(ClO\u2083)\u2082.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of Mn(ClO\u2083)\u2082 = 0.274 g<\/li>\n\n\n\n<li>Moles of Mn(ClO\u2083)\u2082 = mass \/ molar mass<br>Moles of Mn(ClO\u2083)\u2082 = 0.274 g \/ 221.84 g\/mol<br>Moles of Mn(ClO\u2083)\u2082 = 1.236 \u00d7 10\u207b\u2074 mol<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 3: Calculate the molarity.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume of solution = 100.0 mL = 0.1000 L<\/li>\n\n\n\n<li>Molarity (M) = moles of solute \/ volume of solution (in liters)<br>Molarity (M) = 1.236 \u00d7 10\u207b\u2074 mol \/ 0.1000 L<br>Molarity (M) = 1.236 \u00d7 10\u207b\u00b3 M<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">(b) Mass Percent of Mn(ClO\u2083)\u2082 in solution<\/h3>\n\n\n\n<p><strong>Step 1: Calculate the total mass of the solution.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of Mn(ClO\u2083)\u2082 = 0.274 g<\/li>\n\n\n\n<li>Mass of water = 10.00 g<br>Total mass of solution = 0.274 g + 10.00 g = 10.274 g<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 2: Calculate the mass percent.<\/strong><br>Mass percent = (mass of solute \/ total mass of solution) \u00d7 100%<br>Mass percent = (0.274 g \/ 10.274 g) \u00d7 100%<br>Mass percent \u2248 2.67%<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answers:<\/h3>\n\n\n\n<p>(a) The molarity of manganese chlorate in solution is <strong>1.236 \u00d7 10\u207b\u00b3 M<\/strong>.<br>(b) The mass percent of manganese chlorate in the solution is <strong>2.67%<\/strong>.<\/p>\n\n\n\n<p>This method ensures precise and reliable calculations, using basic principles of molar mass, moles, and mass percent.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner9-214.jpeg\" alt=\"\" class=\"wp-image-258691\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>0.274 g manganese chlorate (Mn(ClO3)2) is dissolved in water to make 100.0 mL of solution. Calculate the molarity of manganese chlorate in solution. (b) Calculate the mass percent of manganese chlorate in a solution made by dissolving 0.274 g Mn(ClO3)2 in 10.00 g water. The Correct Answer and Explanation is: (a) Molarity of manganese chlorate [&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-258690","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/258690","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=258690"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/258690\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=258690"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=258690"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=258690"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}