{"id":204867,"date":"2025-03-22T17:42:37","date_gmt":"2025-03-22T17:42:37","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=204867"},"modified":"2025-03-22T17:42:40","modified_gmt":"2025-03-22T17:42:40","slug":"what-mass-of-potassium-hydroxide-koh-molar-mass-56-1-g-mol","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/22\/what-mass-of-potassium-hydroxide-koh-molar-mass-56-1-g-mol\/","title":{"rendered":"What Mass Of Potassium Hydroxide (KOH, Molar Mass &#8211; 56.1 G\/Mol)"},"content":{"rendered":"\n<p>What Mass Of Potassium Hydroxide (KOH, Molar Mass &#8211; 56.1 G\/Mol) Is Needed To Make 100.0 ML Of A 0.125 M KOH Solution? O 7.01 G 2238 O 0.7018 O 0,07018<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\">The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Correct Answer: 0.701 g<\/strong><\/h3>\n\n\n\n<p>To determine the mass of potassium hydroxide (KOH) required to prepare a 100.0 mL (0.100 L) solution with a concentration of <strong>0.125 M<\/strong>, we use the formula: Mass=Molarity\u00d7Volume\u00d7Molar&nbsp;Mass\\text{Mass} = \\text{Molarity} \\times \\text{Volume} \\times \\text{Molar Mass}<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Molarity (M)<\/strong> = 0.125 M<\/li>\n\n\n\n<li><strong>Volume (V)<\/strong> = 100.0 mL = 0.100 L<\/li>\n\n\n\n<li><strong>Molar Mass of KOH<\/strong> = 56.1 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>Substituting the values: Mass=(0.125&nbsp;mol\/L)\u00d7(0.100&nbsp;L)\u00d7(56.1&nbsp;g\/mol)\\text{Mass} = (0.125 \\text{ mol\/L}) \\times (0.100 \\text{ L}) \\times (56.1 \\text{ g\/mol}) Mass=0.701&nbsp;g\\text{Mass} = 0.701 \\text{ g}<\/p>\n\n\n\n<p>Thus, the correct answer is <strong>0.701 g (rounded from 0.70125 g).<\/strong><\/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<ol class=\"wp-block-list\">\n<li><strong>Understanding Molarity:<\/strong><br>Molarity (M) is the number of moles of solute per liter of solution. A <strong>0.125 M<\/strong> solution means there are <strong>0.125 moles of KOH per 1 liter of solution<\/strong>.<\/li>\n\n\n\n<li><strong>Converting Volume:<\/strong><br>The problem gives volume in milliliters (<strong>100.0 mL<\/strong>), but molarity uses liters, so we convert: 100.0\u00a0mL=0.100\u00a0L100.0 \\text{ mL} = 0.100 \\text{ L}<\/li>\n\n\n\n<li><strong>Finding the Number of Moles of KOH:<\/strong><br>Using the definition of molarity: Moles\u00a0of\u00a0KOH=Molarity\u00d7Volume\\text{Moles of KOH} = \\text{Molarity} \\times \\text{Volume} =(0.125\u00a0mol\/L)\u00d7(0.100\u00a0L)= (0.125 \\text{ mol\/L}) \\times (0.100 \\text{ L}) =0.0125\u00a0moles= 0.0125 \\text{ moles}<\/li>\n\n\n\n<li><strong>Calculating Mass of KOH:<\/strong> Mass=Moles\u00d7Molar\u00a0Mass\\text{Mass} = \\text{Moles} \\times \\text{Molar Mass} =(0.0125\u00a0moles)\u00d7(56.1\u00a0g\/mol)= (0.0125 \\text{ moles}) \\times (56.1 \\text{ g\/mol}) =0.701\u00a0g= 0.701 \\text{ g}<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Conclusion:<\/strong><\/h3>\n\n\n\n<p>To prepare <strong>100.0 mL of 0.125 M KOH solution<\/strong>, you need to weigh <strong>0.701 g of KOH<\/strong> and dissolve it in enough water to make the total volume 100.0 mL.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Generated Image Prompt:<\/strong><\/h3>\n\n\n\n<p><em>A digital illustration of a chemistry lab setup showing a beaker with 100.0 mL of clear KOH solution, a balance weighing 0.701 g of KOH powder, and a labeled molarity equation.<\/em><\/p>\n\n\n\n<p>Generating the image now\u2026<\/p>\n\n\n\n<p>Here is the generated illustration of the chemistry lab setup showing the preparation of the KOH solution. Let me know if you need any modifications or additional details!<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/03\/image-1151.png\" alt=\"\" class=\"wp-image-204868\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>What Mass Of Potassium Hydroxide (KOH, Molar Mass &#8211; 56.1 G\/Mol) Is Needed To Make 100.0 ML Of A 0.125 M KOH Solution? O 7.01 G 2238 O 0.7018 O 0,07018 The correct answer and explanation is: Correct Answer: 0.701 g To determine the mass of potassium hydroxide (KOH) required to prepare a 100.0 mL [&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-204867","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/204867","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=204867"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/204867\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=204867"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=204867"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=204867"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}