What is the molar mass of alumina, Al2O3? The atomic weight of aluminium is 26.98 and the atomic weight of oxygen is 16.00.<\/p>\n\n\n\n
42.98 g\/mol
48.00 g\/mol
149.96 g\/mol
101.96 g\/mol<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n To determine the molar mass of alumina (Al\u2082O\u2083)<\/strong>, we need to calculate the total mass of all the atoms in one mole of the compound.<\/p>\n\n\n\n Chemical formula of alumina<\/strong>: Al\u2082O\u2083 Atomic weights (given)<\/strong>:<\/p>\n\n\n\n Now, calculate the molar mass: Molar mass of Al2O3=(2\u00d726.98)+(3\u00d716.00)\\text{Molar mass of Al}_2\\text{O}_3 = (2 \\times 26.98) + (3 \\times 16.00) =53.96+48.00=101.96\u2009g\/mol= 53.96 + 48.00 = \\boxed{101.96 \\, \\text{g\/mol}}<\/p>\n\n\n\n The molar mass<\/strong> of a compound refers to the mass of one mole of its molecules. A mole is a standard scientific unit for measuring large quantities of very small entities such as atoms or molecules. One mole of any substance contains Avogadro\u2019s number of particles<\/strong>, which is 6.022\u00d710236.022 \\times 10^{23}.<\/p>\n\n\n\n Alumina, or aluminum oxide (Al\u2082O\u2083)<\/strong>, is a white crystalline compound commonly found in nature as the mineral corundum<\/strong>. It is a significant raw material in the production of aluminum metal and also has applications in ceramics and refractories due to its hardness and high melting point.<\/p>\n\n\n\n When calculating the molar mass of a compound, each element\u2019s atomic weight is multiplied by the number of times that element appears in the formula, and the products are then added together.<\/p>\n\n\n\n In this case:<\/p>\n\n\n\n Adding these gives a total molar mass of 101.96 g\/mol<\/strong>, which is the correct and final answer.<\/p>\n\n\n\n What is the molar mass of alumina, Al2O3? The atomic weight of aluminium is 26.98 and the atomic weight of oxygen is 16.00. 42.98 g\/mol48.00 g\/mol149.96 g\/mol101.96 g\/mol The Correct Answer and Explanation is: To determine the molar mass of alumina (Al\u2082O\u2083), we need to calculate the total mass of all the atoms in one […]<\/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-213134","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/213134","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=213134"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/213134\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=213134"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=213134"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=213134"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step-by-Step Calculation:<\/h3>\n\n\n\n
This means the compound contains:<\/p>\n\n\n\n\n
\n
\n\n\n\nExplanation:<\/h3>\n\n\n\n
\n
\u2705 Correct Answer: 101.96 g\/<\/strong><\/h3>\n","protected":false},"excerpt":{"rendered":"