{"id":249778,"date":"2025-07-09T21:09:49","date_gmt":"2025-07-09T21:09:49","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=249778"},"modified":"2025-07-09T21:09:51","modified_gmt":"2025-07-09T21:09:51","slug":"how-many-grams-of-naoh-are-required-to-prepare-240-ml-of-a-3-50-m-naoh-solution","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/09\/how-many-grams-of-naoh-are-required-to-prepare-240-ml-of-a-3-50-m-naoh-solution\/","title":{"rendered":"How many grams of NaOH are required to prepare 240 mL of a 3.50 M NaOH solution"},"content":{"rendered":"\n<p>How many grams of NaOH are required to prepare 240 mL of a 3.50 M NaOH solution? How many grams of NaOH are required to prepare 240 mL of a 3.50 M NaOH solution? 0.840 g 33.6 g 54.8 g 583 g<\/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<p>To solve this, we will use the relationship between molarity, volume, and moles, which is given by the formula:Molarity&nbsp;(M)=moles&nbsp;of&nbsp;solutevolume&nbsp;of&nbsp;solution&nbsp;in&nbsp;liters\\text{Molarity (M)} = \\frac{\\text{moles of solute}}{\\text{volume of solution in liters}}Molarity&nbsp;(M)=volume&nbsp;of&nbsp;solution&nbsp;in&nbsp;litersmoles&nbsp;of&nbsp;solute\u200b<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Find the moles of NaOH needed.<\/h3>\n\n\n\n<p>We know the molarity of the NaOH solution is 3.50 M (moles per liter), and the volume required is 240 mL. First, we convert the volume from milliliters to liters:240\u2009mL=0.240\u2009L240 \\, \\text{mL} = 0.240 \\, \\text{L}240mL=0.240L<\/p>\n\n\n\n<p>Now, use the molarity formula to find the moles of NaOH:moles&nbsp;of&nbsp;NaOH=Molarity\u00d7Volume\\text{moles of NaOH} = \\text{Molarity} \\times \\text{Volume}moles&nbsp;of&nbsp;NaOH=Molarity\u00d7Volumemoles&nbsp;of&nbsp;NaOH=3.50\u2009M\u00d70.240\u2009L\\text{moles of NaOH} = 3.50 \\, \\text{M} \\times 0.240 \\, \\text{L}moles&nbsp;of&nbsp;NaOH=3.50M\u00d70.240Lmoles&nbsp;of&nbsp;NaOH=0.84\u2009moles\\text{moles of NaOH} = 0.84 \\, \\text{moles}moles&nbsp;of&nbsp;NaOH=0.84moles<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Convert moles to grams.<\/h3>\n\n\n\n<p>Now, to convert moles of NaOH to grams, we need the molar mass of NaOH. The molar mass is the sum of the atomic masses of sodium (Na), oxygen (O), and hydrogen (H):Molar&nbsp;mass&nbsp;of&nbsp;NaOH=22.99\u2009g\/mol&nbsp;(Na)+15.999\u2009g\/mol&nbsp;(O)+1.008\u2009g\/mol&nbsp;(H)=39.997\u2009g\/mol\\text{Molar mass of NaOH} = 22.99 \\, \\text{g\/mol (Na)} + 15.999 \\, \\text{g\/mol (O)} + 1.008 \\, \\text{g\/mol (H)} = 39.997 \\, \\text{g\/mol}Molar&nbsp;mass&nbsp;of&nbsp;NaOH=22.99g\/mol&nbsp;(Na)+15.999g\/mol&nbsp;(O)+1.008g\/mol&nbsp;(H)=39.997g\/mol<\/p>\n\n\n\n<p>Now, calculate the mass in grams:Mass&nbsp;of&nbsp;NaOH=moles&nbsp;of&nbsp;NaOH\u00d7molar&nbsp;mass&nbsp;of&nbsp;NaOH\\text{Mass of NaOH} = \\text{moles of NaOH} \\times \\text{molar mass of NaOH}Mass&nbsp;of&nbsp;NaOH=moles&nbsp;of&nbsp;NaOH\u00d7molar&nbsp;mass&nbsp;of&nbsp;NaOHMass&nbsp;of&nbsp;NaOH=0.84\u2009moles\u00d739.997\u2009g\/mol\\text{Mass of NaOH} = 0.84 \\, \\text{moles} \\times 39.997 \\, \\text{g\/mol}Mass&nbsp;of&nbsp;NaOH=0.84moles\u00d739.997g\/molMass&nbsp;of&nbsp;NaOH\u224833.6\u2009g\\text{Mass of NaOH} \\approx 33.6 \\, \\text{g}Mass&nbsp;of&nbsp;NaOH\u224833.6g<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Conclusion.<\/h3>\n\n\n\n<p>The correct answer is <strong>33.6 grams<\/strong> of NaOH required to prepare 240 mL of a 3.50 M NaOH solution.<\/p>\n\n\n\n<p>Thus, the answer is <strong>33.6 g<\/strong>.<\/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-banner6-53.jpeg\" alt=\"\" class=\"wp-image-249779\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>How many grams of NaOH are required to prepare 240 mL of a 3.50 M NaOH solution? How many grams of NaOH are required to prepare 240 mL of a 3.50 M NaOH solution? 0.840 g 33.6 g 54.8 g 583 g The Correct Answer and Explanation is: To solve this, we will use the [&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-249778","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/249778","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=249778"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/249778\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=249778"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=249778"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=249778"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}