{"id":222405,"date":"2025-05-31T10:09:25","date_gmt":"2025-05-31T10:09:25","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=222405"},"modified":"2025-05-31T10:09:27","modified_gmt":"2025-05-31T10:09:27","slug":"a-chemist-prepares-a-solution-of-vanadium","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/31\/a-chemist-prepares-a-solution-of-vanadium\/","title":{"rendered":"A chemist prepares a solution of vanadium"},"content":{"rendered":"\n

A chemist prepares a solution of vanadium<\/p>\n\n\n\n

III bromide \u00c3\u201a\u00c2 VBr3 \u00c3\u201a\u00c2 by measuring out 0.75g of VBr3 into a 250.mL volumetric flask and filling to the mark with distilled water.
\u00c3\u201a\u00c2<\/p>\n\n\n\n

Calculate the molarity of<\/p>\n\n\n\n

Br\u00c3\u00a2\u00cb\u2020\u00e2\u20ac\u2122 anions in the chemist’s solution.
\u00c3\u201a\u00c2<\/p>\n\n\n\n

Be sure your answer is rounded to the correct number of significant digits.
molL<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong>To calculate the molarity of Br\u207b anions<\/strong> in the chemist’s solution, we follow these steps:<\/p>\n\n\n\n

\n\n\n\n

Step 1: Determine the molar mass of VBr\u2083<\/strong><\/h3>\n\n\n\n

Vanadium (V): 50.94 g\/mol
Bromine (Br): 79.90 g\/mol \u00d7 3 = 239.70 g\/mol
Total molar mass of VBr\u2083 = 50.94 + 239.70 = 290.64 g\/mol<\/strong><\/p>\n\n\n\n

\n\n\n\n

Step 2: Calculate the moles of VBr\u2083 used<\/strong><\/h3>\n\n\n\n

Given mass of VBr\u2083 = 0.75 g
Moles of VBr\u2083 = 0.75 g290.64 g\/mol=0.00258 mol\\frac{0.75\\ \\text{g}}{290.64\\ \\text{g\/mol}} = 0.00258\\ \\text{mol}<\/p>\n\n\n\n

\n\n\n\n

Step 3: Determine the moles of Br\u207b anions<\/strong><\/h3>\n\n\n\n

Each formula unit of VBr\u2083 contains 3 Br\u207b ions<\/strong>, so: Moles of Br\u207b=0.00258 mol VBr\u2083\u00d73=0.00774 mol Br\u207b\\text{Moles of Br\u207b} = 0.00258\\ \\text{mol VBr\u2083} \u00d7 3 = 0.00774\\ \\text{mol Br\u207b}<\/p>\n\n\n\n

\n\n\n\n

Step 4: Calculate the molarity of Br\u207b<\/strong><\/h3>\n\n\n\n

Volume of solution = 250. mL = 0.250 L Molarity (M)=0.00774 mol Br\u207b0.250 L=0.03096 mol\/L\\text{Molarity (M)} = \\frac{0.00774\\ \\text{mol Br\u207b}}{0.250\\ \\text{L}} = 0.03096\\ \\text{mol\/L}<\/p>\n\n\n\n

Rounded to 3 significant digits<\/strong> (since 0.75 g has 3 sig. figs):<\/p>\n\n\n\n

\u2705 Final Answer: 0.03096 mol\/L Br\u207b<\/strong><\/h3>\n\n\n\n
\n\n\n\n

\ud83d\udca1 Explanation<\/strong><\/h3>\n\n\n\n

In this problem, we are asked to find the molarity of bromide ions (Br\u207b)<\/strong> in a solution made by dissolving a specific mass of vanadium(III) bromide (VBr\u2083)<\/strong> in water. Molarity refers to the number of moles of a solute per liter of solution, expressed in mol\/L. Since VBr\u2083 is an ionic compound, it dissociates completely in water into vanadium ions and bromide ions. Specifically, each mole of VBr\u2083 releases three moles of Br\u207b<\/strong> ions because its chemical formula contains three bromine atoms.<\/p>\n\n\n\n

The process starts by determining the molar mass<\/strong> of VBr\u2083, which is the sum of the atomic masses of vanadium and three bromines. After that, we calculate the moles of VBr\u2083<\/strong> by dividing the mass used (0.75 g) by the molar mass. From there, we multiply by 3 to find the moles of bromide ions, because each VBr\u2083 unit produces 3 Br\u207b.<\/p>\n\n\n\n

Finally, the moles of Br\u207b are divided by the total solution volume in liters (250 mL = 0.250 L) to obtain the molarity of Br\u207b ions<\/strong>. The answer is rounded to three significant digits, matching the precision of the given mass (0.75 g).<\/p>\n\n\n\n

This calculation is important in chemistry, particularly in preparing solutions for reactions, titrations, or analytical procedures where knowing ion concentration is essential.<\/p>\n\n\n\n

\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

A chemist prepares a solution of vanadium III bromide \u00c3\u201a\u00c2 VBr3 \u00c3\u201a\u00c2 by measuring out 0.75g of VBr3 into a 250.mL volumetric flask and filling to the mark with distilled water.\u00c3\u201a\u00c2 Calculate the molarity of Br\u00c3\u00a2\u00cb\u2020\u00e2\u20ac\u2122 anions in the chemist’s solution.\u00c3\u201a\u00c2 Be sure your answer is rounded to the correct number of significant digits.molL The […]<\/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-222405","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222405","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=222405"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222405\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=222405"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=222405"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=222405"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}