{"id":223528,"date":"2025-06-02T03:38:49","date_gmt":"2025-06-02T03:38:49","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=223528"},"modified":"2025-06-02T03:38:51","modified_gmt":"2025-06-02T03:38:51","slug":"potassium-iodate-is-214-001-g-mol","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/02\/potassium-iodate-is-214-001-g-mol\/","title":{"rendered":"potassium iodate is 214.001 g\/mol"},"content":{"rendered":"\n<p>potassium iodate is 214.001 g\/mol. Mass of KIO3: 0.630 g M = ((mass KIO3 (g) \/ molar mass KIO3) \/ 0.250 L)<br>Calculate the molarity of the solution. The volume of the standard solution is 250 mL. The molar mass of potassium iodate is 214.001 g\/mol. Mass of KIO3: 0.630 g M = ((mass KIO3 (g) \/ molar mass KIO3) \/ 0.250 L)<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To calculate the molarity (M) of the potassium iodate (KIO\u2083) solution:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Given:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of KIO\u2083 = 214.001 g\/mol<\/li>\n\n\n\n<li>Mass of KIO\u2083 = 0.630 g<\/li>\n\n\n\n<li>Volume of solution = 250 mL = 0.250 L<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step-by-step Calculation:<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Calculate the number of moles of KIO\u2083<\/strong><\/li>\n<\/ol>\n\n\n\n<p>Moles&nbsp;of&nbsp;KIO\u2083=MassMolar&nbsp;Mass=0.630&nbsp;g214.001&nbsp;g\/mol=0.002945&nbsp;mol\\text{Moles of KIO\u2083} = \\frac{\\text{Mass}}{\\text{Molar Mass}} = \\frac{0.630\\ \\text{g}}{214.001\\ \\text{g\/mol}} = 0.002945\\ \\text{mol}<\/p>\n\n\n\n<ol start=\"2\" class=\"wp-block-list\">\n<li><strong>Calculate the molarity (M)<\/strong><\/li>\n<\/ol>\n\n\n\n<p>M=moles&nbsp;of&nbsp;solutevolume&nbsp;of&nbsp;solution&nbsp;in&nbsp;liters=0.0029450.250=0.01178&nbsp;mol\/LM = \\frac{\\text{moles of solute}}{\\text{volume of solution in liters}} = \\frac{0.002945}{0.250} = 0.01178\\ \\text{mol\/L}<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Final Answer:<\/strong><\/h3>\n\n\n\n<p>0.01178&nbsp;M\\boxed{0.01178\\ \\text{M}}<\/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<p>Molarity is a measure of the concentration of a solute in a solution. It is expressed in moles of solute per liter of solution (mol\/L). In this problem, we are asked to find the molarity of a potassium iodate (KIO\u2083) solution, given its mass and the final volume of the solution.<\/p>\n\n\n\n<p>First, we convert the given mass of KIO\u2083 into moles using its molar mass. The molar mass represents how much one mole of a substance weighs. Potassium iodate (KIO\u2083) has a molar mass of 214.001 g\/mol. By dividing the mass (0.630 g) by this molar mass, we determine the number of moles of KIO\u2083 present in the solution, which turns out to be approximately 0.002945 mol.<\/p>\n\n\n\n<p>Next, we convert the given volume into liters, since molarity requires the volume in liters. The problem states that the volume is 250 mL, which is equal to 0.250 L. Now, using the definition of molarity (moles of solute divided by liters of solution), we divide the number of moles by the volume in liters. This yields a molarity of 0.01178 mol\/L.<\/p>\n\n\n\n<p>This calculation is useful in chemistry for preparing standard solutions, performing titrations, or determining reaction stoichiometry. Accurate molarity is crucial for experimental precision. The steps followed\u2014calculating moles, converting volume, and applying the molarity formula\u2014are foundational for quantitative solution preparation.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner4-91.jpeg\" alt=\"\" class=\"wp-image-223529\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>potassium iodate is 214.001 g\/mol. Mass of KIO3: 0.630 g M = ((mass KIO3 (g) \/ molar mass KIO3) \/ 0.250 L)Calculate the molarity of the solution. The volume of the standard solution is 250 mL. The molar mass of potassium iodate is 214.001 g\/mol. Mass of KIO3: 0.630 g M = ((mass KIO3 (g) [&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-223528","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223528","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=223528"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223528\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=223528"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=223528"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=223528"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}