{"id":20549,"date":"2025-06-14T14:18:07","date_gmt":"2025-06-14T14:18:07","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=20549"},"modified":"2025-06-14T14:18:08","modified_gmt":"2025-06-14T14:18:08","slug":"you-require-a-solution-containing-1-15-mm-of-sodium-hydroxide","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/you-require-a-solution-containing-1-15-mm-of-sodium-hydroxide\/","title":{"rendered":"You require a solution containing 1.15 mM of sodium hydroxide"},"content":{"rendered":"\n<p>You require a solution containing 1.15 mM of sodium hydroxide. What is the maximum volume of solution (L) that can be prepared from 0.27 kg of sodium hydroxide (whole number)?<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\"><strong>The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 <strong>Correct Answer:<\/strong> 5,870 L<\/h3>\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>To determine the <strong>maximum volume<\/strong> of a <strong>1.15 mM sodium hydroxide (NaOH)<\/strong> solution that can be prepared from <strong>0.27 kg of NaOH<\/strong>, follow these steps:<\/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: Convert mass to moles<\/strong><\/h3>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of NaOH = <strong>0.27 kg<\/strong> = <strong>270 g<\/strong><\/li>\n\n\n\n<li>Molar mass of NaOH = <strong>22.99 (Na) + 15.999 (O) + 1.008 (H) = 39.997 g\/mol \u2248 40 g\/mol<\/strong><\/li>\n<\/ul>\n\n\n\n<p>Moles&nbsp;of&nbsp;NaOH=MassMolar&nbsp;mass=270&nbsp;g40&nbsp;g\/mol=6.75&nbsp;mol\\text{Moles of NaOH} = \\frac{\\text{Mass}}{\\text{Molar mass}} = \\frac{270\\text{ g}}{40\\text{ g\/mol}} = 6.75\\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 molarity formula<\/strong><\/h3>\n\n\n\n<p>Molarity (M) = moles \/ volume (L)<\/p>\n\n\n\n<p>We rearrange to find volume: Volume&nbsp;(L)=MolesMolarity=6.75&nbsp;mol1.15\u00d710\u22123&nbsp;mol\/L=6.750.00115\u22485869.565&nbsp;L\\text{Volume (L)} = \\frac{\\text{Moles}}{\\text{Molarity}} = \\frac{6.75\\ \\text{mol}}{1.15 \\times 10^{-3}\\ \\text{mol\/L}} = \\frac{6.75}{0.00115} \\approx 5869.565 \\text{ L}<\/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 3: Round to whole number<\/strong><\/h3>\n\n\n\n<p>Since the question asks for the <strong>maximum volume in whole liters<\/strong>: 5,870&nbsp;L\\boxed{5,870\\ \\text{L}}<\/p>\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>From 0.27 kg of sodium hydroxide, you can prepare a <strong>maximum of 5,870 liters<\/strong> of a <strong>1.15 mM NaOH<\/strong> solution. This calculation is vital in chemical solution preparation where concentration accuracy and resource efficiency are critical in labs, industries, and education.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>You require a solution containing 1.15 mM of sodium hydroxide. What is the maximum volume of solution (L) that can be prepared from 0.27 kg of sodium hydroxide (whole number)? The correct answer and explanation is: \u2705 Correct Answer: 5,870 L Explanation: To determine the maximum volume of a 1.15 mM sodium hydroxide (NaOH) solution [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-20549","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/20549","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=20549"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/20549\/revisions"}],"predecessor-version":[{"id":20550,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/20549\/revisions\/20550"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=20549"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=20549"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=20549"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}