{"id":5132,"date":"2019-02-17T22:15:18","date_gmt":"2019-02-17T16:45:18","guid":{"rendered":"https:\/\/physicscatalyst.com\/article\/?p=5132"},"modified":"2024-08-12T05:10:55","modified_gmt":"2024-08-11T23:40:55","slug":"molarity-practice-worksheet","status":"publish","type":"post","link":"https:\/\/physicscatalyst.com\/article\/molarity-practice-worksheet\/","title":{"rendered":"Molarity Practice Worksheet"},"content":{"rendered":"\n<p><strong>Question 1<\/strong><\/p>\n\n\n\n<p>Calculate the <a href=\"https:\/\/physicscatalyst.com\/chemistry\/molarity-formula.php\" rel=\"noopener noreferrer\">Molarity<\/a> of the below solutions<\/p>\n\n\n\n<p>a.100 g NaCl dissolved in 1500 cm<sup>3<\/sup> solution<\/p>\n\n\n\n<p>b..2 Moles of AgNO<sub>3<\/sub> in 2 L of solution<\/p>\n\n\n\n<p>c.16 gm of CH<sub>3<\/sub>OH in 400 cm<sup>3<\/sup> of solution<\/p>\n\n\n\n<p>Molar Mass of NaCL = 58.5 g\/Moles<\/p>\n\n\n\n<p>Molar Mass of CH<sub>3<\/sub> OH = 32 g\/Moles<\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>a.14 M<\/p>\n\n\n\n<p>b..1 M<\/p>\n\n\n\n<p>c. 25 M<\/p>\n\n\n\n<p><strong>Question 2<\/strong><\/p>\n\n\n\n<p>Find the volume of the 2.5 M sugar solution which contains .2 moles of sugar?<\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>.08 L<\/p>\n\n\n\n<p><strong>Question 3<\/strong><\/p>\n\n\n\n<p>How much solute A is required to prepare id 400 ml of M\/4 solution? Given Molar Mass of A = 132.1 g\/M<\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>13.21 g<\/p>\n\n\n\n<p><strong>Question 4<\/strong><\/p>\n\n\n\n<p>An aqueous solution of the substance A contain 28%&nbsp; of A by weight. The density of the solution is 1.2 g\/mL. Find the molarity of the solution<\/p>\n\n\n\n<p>Given the Molar Mass of the substance A = 36.5 g\/Moles<\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>9.2 M<\/p>\n\n\n\n<p><strong>Question 5.<\/strong><\/p>\n\n\n\n<p>A solution is prepared by mixing 2L of 4 M HCL solution and 3L of 1.5 M HCL solution. Find the Molarity of the final solution?<\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>2.5 M<\/p>\n\n\n\n<p><strong>Question 6<\/strong><\/p>\n\n\n\n<p>A solution of 3 L and 1 M NaOH is prepared by mixing solutions of 2.50 M and .400 M NaOH respectively. Find the volume of the solution added<\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>.857 L of 2.5 M<\/p>\n\n\n\n<p>2.14 L of .4 M<\/p>\n\n\n\n<p><strong>Question 7<\/strong><\/p>\n\n\n\n<p>We have chemical equation given as<\/p>\n\n\n\n<p>Zn + H<sub>2<\/sub> SO<sub>4<\/sub>&nbsp; -&gt; Zn SO<sub>4<\/sub> + H<sub>2<\/sub><\/p>\n\n\n\n<p>Calculate the volume of 2 M H<sub>2<\/sub> SO<sub>4<\/sub> required to react with 10g of Zinc<\/p>\n\n\n\n<p>Molar Mass of Zn= 65.4 g\/Moles<\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>76.5 mL<\/p>\n\n\n\n<p><strong>Question 8<\/strong><\/p>\n\n\n\n<p>How much water should be added to 5L of 6.0 M HCL to make a solution of 2.0M HCL?<\/p>\n\n\n\n<p><strong>Answer<\/strong><\/p>\n\n\n\n<p>10 L<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Question 1 Calculate the Molarity of the below solutions a.100 g NaCl dissolved in 1500 cm3 solution b..2 Moles of AgNO3 in 2 L of solution c.16 gm of CH3OH in 400 cm3 of solution Molar Mass of NaCL = 58.5 g\/Moles Molar Mass of CH3 OH = 32 g\/Moles Answer a.14 M b..1 M [&hellip;]<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_uag_custom_page_level_css":"","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":"","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":"set","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":[499],"tags":[],"class_list":["post-5132","post","type-post","status-publish","format-standard","hentry","category-chemistry"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.3 - 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