{"id":36042,"date":"2025-06-25T01:52:52","date_gmt":"2025-06-25T01:52:52","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=36042"},"modified":"2025-06-25T01:52:53","modified_gmt":"2025-06-25T01:52:53","slug":"what-is-the-molar-mass-of-nano3","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/what-is-the-molar-mass-of-nano3\/","title":{"rendered":"What is the molar mass of NaNO3"},"content":{"rendered":"\n<p>What is the molar mass of NaNO3? A. 34.0 g B. 37.0 g C. 53.0 g D. 75.0 g E. 85.0 g<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>The correct answer is:<br><strong>E. 85.0 g<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>To find the <strong>molar mass<\/strong> of a compound like sodium nitrate (NaNO\u2083), you must add up the atomic masses of each element present in one mole of the compound. Use the periodic table values for each atom:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Sodium (Na)<\/strong>: 22.99 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, and there are <strong>three<\/strong> oxygen atoms<\/li>\n<\/ul>\n\n\n\n<p>Now calculate the total molar mass:Molar&nbsp;mass&nbsp;of&nbsp;NaNO3=(1\u00d722.99)+(1\u00d714.01)+(3\u00d716.00)\\text{Molar mass of NaNO}_3 = (1 \\times 22.99) + (1 \\times 14.01) + (3 \\times 16.00)Molar&nbsp;mass&nbsp;of&nbsp;NaNO3\u200b=(1\u00d722.99)+(1\u00d714.01)+(3\u00d716.00)=22.99+14.01+48.00=85.00&nbsp;g\/mol= 22.99 + 14.01 + 48.00 = 85.00 \\text{ g\/mol}=22.99+14.01+48.00=85.00&nbsp;g\/mol<\/p>\n\n\n\n<p>Thus, the molar mass of sodium nitrate is <strong>85.0 grams per mole<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Why It Matters:<\/h3>\n\n\n\n<p>Understanding molar mass is critical in chemistry because it allows us to convert between the mass of a substance and the number of moles. Moles are a standard unit in chemistry for counting particles like atoms or molecules.<\/p>\n\n\n\n<p>For example, if you are asked to prepare a solution with one mole of NaNO\u2083, you would weigh out exactly 85.0 grams. This ensures proper stoichiometry in reactions, meaning the chemical quantities react in correct proportions.<\/p>\n\n\n\n<p>This calculation is also essential in fields like environmental science, where sodium nitrate may be a component in fertilizers, and in medical laboratories where precise chemical measurements are required for accurate testing and analysis.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-737.jpeg\" alt=\"\" class=\"wp-image-36043\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-737.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-737-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-737-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>What is the molar mass of NaNO3? A. 34.0 g B. 37.0 g C. 53.0 g D. 75.0 g E. 85.0 g The Correct Answer and Explanation is: The correct answer is:E. 85.0 g Explanation: To find the molar mass of a compound like sodium nitrate (NaNO\u2083), you must add up the atomic masses of [&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-36042","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36042","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=36042"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36042\/revisions"}],"predecessor-version":[{"id":36044,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36042\/revisions\/36044"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=36042"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=36042"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=36042"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}