Molar mass of KAl(SO4)^2\u00b712H2O<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n To calculate the molar mass of potassium aluminum sulfate dodecahydrate, commonly referred to as KAl(SO\u2084)\u2082\u00b712H\u2082O, we first need to determine the molar masses of each individual component in the compound.<\/p>\n\n\n\n Using the periodic table:<\/p>\n\n\n\n Adding all these values together gives us the total molar mass of KAl(SO\u2084)\u2082\u00b712H\u2082O:<\/p>\n\n\n\n [ Thus, the molar mass of potassium aluminum sulfate dodecahydrate, KAl(SO\u2084)\u2082\u00b712H\u2082O, is 474.46 g\/mol<\/strong>. This compound is often utilized in various applications such as water purification, as a food additive, and in the textile industry. Understanding its molar mass is essential for stoichiometric calculations in chemical reactions, as it enables chemists to convert between grams and moles effectively, allowing for precise measurements and formulations in laboratory settings.<\/p>\n","protected":false},"excerpt":{"rendered":" Molar mass of KAl(SO4)^2\u00b712H2O The Correct Answer and Explanation is: To calculate the molar mass of potassium aluminum sulfate dodecahydrate, commonly referred to as KAl(SO\u2084)\u2082\u00b712H\u2082O, we first need to determine the molar masses of each individual component in the compound. Step 1: Identify the Components Step 2: Molar Mass of Each Element Using the periodic […]<\/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-157361","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/157361","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=157361"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/157361\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=157361"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=157361"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=157361"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Identify the Components<\/h3>\n\n\n\n
\n
Step 2: Molar Mass of Each Element<\/h3>\n\n\n\n
\n
Step 3: Calculate Molar Mass of Each Component<\/h3>\n\n\n\n
\n
(1 \\times 39.10 = 39.10 \\, \\text{g\/mol})<\/li>\n\n\n\n
(1 \\times 26.98 = 26.98 \\, \\text{g\/mol})<\/li>\n\n\n\n
(2 \\times 32.07 = 64.14 \\, \\text{g\/mol})<\/li>\n\n\n\n
(8 \\times 16.00 = 128.00 \\, \\text{g\/mol})<\/li>\n\n\n\n
[
12 \\times (2 \\times 1.01 + 16.00) = 12 \\times 18.02 = 216.24 \\, \\text{g\/mol}
]<\/li>\n<\/ol>\n\n\n\nStep 4: Total Molar Mass Calculation<\/h3>\n\n\n\n
39.10 + 26.98 + 64.14 + 128.00 + 216.24 = 474.46 \\, \\text{g\/mol}
]<\/p>\n\n\n\nConclusion<\/h3>\n\n\n\n