{"id":226415,"date":"2025-06-05T08:49:40","date_gmt":"2025-06-05T08:49:40","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=226415"},"modified":"2025-06-05T08:49:42","modified_gmt":"2025-06-05T08:49:42","slug":"what-is-the-molar-mass-of-silver-nitrate","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/05\/what-is-the-molar-mass-of-silver-nitrate\/","title":{"rendered":"What is the molar mass of silver nitrate"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">What is the molar mass of silver nitrate ?<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><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 class=\"wp-block-paragraph\"><strong>Correct Answer:<\/strong><br>The molar mass of <strong>silver nitrate (AgNO\u2083)<\/strong> is <strong>169.87 g\/mol<\/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<p class=\"wp-block-paragraph\">The molar mass of a compound is the mass of one mole of that substance, expressed in grams per mole (g\/mol). It is calculated by summing the atomic masses of all atoms present in the chemical formula.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Silver nitrate has the chemical formula <strong>AgNO\u2083<\/strong>, meaning it consists of:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>1 silver atom (Ag)<\/strong><\/li>\n\n\n\n<li><strong>1 nitrogen atom (N)<\/strong><\/li>\n\n\n\n<li><strong>3 oxygen atoms (O)<\/strong><\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">To determine the molar mass, we use the atomic masses from the periodic table:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Silver (Ag)<\/strong> = 107.87 g\/mol<\/li>\n\n\n\n<li><strong>Nitrogen (N)<\/strong> = 14.01 g\/mol<\/li>\n\n\n\n<li><strong>Oxygen (O)<\/strong> = 16.00 g\/mol<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Now, we calculate: Molar&nbsp;Mass&nbsp;of&nbsp;AgNO\u2083=(1\u00d7107.87)+(1\u00d714.01)+(3\u00d716.00)\\text{Molar Mass of AgNO\u2083} = (1 \\times 107.87) + (1 \\times 14.01) + (3 \\times 16.00) =107.87+14.01+48.00=169.88&nbsp;g\/mol= 107.87 + 14.01 + 48.00 = \\boxed{169.88\\ \\text{g\/mol}}<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(You may also see 169.87 g\/mol depending on rounding conventions.)<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Importance of Molar Mass:<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Knowing the molar mass of silver nitrate is essential in chemistry for:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Stoichiometry<\/strong> \u2013 determining how much of a substance is needed or produced in a chemical reaction.<\/li>\n\n\n\n<li><strong>Concentration Calculations<\/strong> \u2013 converting between grams and moles when preparing solutions.<\/li>\n\n\n\n<li><strong>Quantitative Analysis<\/strong> \u2013 calculating yields, limiting reagents, or titrations in labs.<\/li>\n<\/ol>\n\n\n\n<p class=\"wp-block-paragraph\">Silver nitrate is a common reagent used in analytical chemistry, especially in precipitation reactions and in tests for halide ions. It\u2019s also used in medicine for its antibacterial properties. Therefore, accurate molar mass is crucial in practical applications where precise measurement is required.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">In summary, the molar mass of <strong>AgNO\u2083<\/strong> is <strong>169.87 g\/mol<\/strong>, calculated by adding the atomic masses of silver, nitrogen, and oxygen atoms.<\/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-banner9-52.jpeg\" alt=\"\" class=\"wp-image-226416\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>What is the molar mass of silver nitrate ? The Correct Answer and Explanation is: Correct Answer:The molar mass of silver nitrate (AgNO\u2083) is 169.87 g\/mol. Explanation The molar mass of a compound is the mass of one mole of that substance, expressed in grams per mole (g\/mol). It is calculated by summing the atomic [&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-226415","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/226415","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=226415"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/226415\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=226415"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=226415"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=226415"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}