How many formula units are contained in 23.5 g of Sb2S3<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n The correct answer is: 4.17 \\times 10^{22}<\/strong><\/mark><\/p>\n\n\n\n To determine the number of formula units in 23.5 g of antimony(III) sulfide (Sb\u2082S\u2083), follow these steps:<\/p>\n\n\n\n First, we need to calculate the molar mass (molecular weight) of Sb\u2082S\u2083. The atomic masses are:<\/p>\n\n\n\n The formula Sb\u2082S\u2083 has:<\/p>\n\n\n\n So, the molar mass of Sb\u2082S\u2083 = (243.52 + 96.18 = 339.70) g\/mol<\/p>\n\n\n\n Using the formula:<\/p>\n\n\n\n [ Substitute the values:<\/p>\n\n\n\n [ One mole of any substance contains (6.022 \\times 10^{23}) formula units (Avogadro’s number). To find the number of formula units in 0.0692 moles:<\/p>\n\n\n\n [ [ In chemistry, a formula unit<\/strong> represents the smallest whole-number ratio of ions in an ionic compound. In this case, antimony(III) sulfide (Sb\u2082S\u2083) consists of two antimony (Sb) atoms and three sulfur (S) atoms. To calculate the number of formula units in a given mass of Sb\u2082S\u2083, we must first know its molar mass, which is the mass of one mole of Sb\u2082S\u2083 formula units. This is calculated by summing the atomic masses of all the atoms present in one formula unit of Sb\u2082S\u2083: two antimony atoms and three sulfur atoms.<\/p>\n\n\n\n Once we have the molar mass, the number of moles present in a given mass (in this case, 23.5 g) can be determined using the formula for moles, which divides the mass by the molar mass. For Sb\u2082S\u2083, we calculated that the molar mass is 339.70 g\/mol, and dividing 23.5 g by this value gives us about 0.0692 moles.<\/p>\n\n\n\n Finally, to find the actual number of formula units, we multiply the number of moles by Avogadro’s number ((6.022 \\times 10^{23})). This gives the total number of Sb\u2082S\u2083 formula units in the sample, which turns out to be approximately (4.17 \\times 10^{22}) formula units. This calculation is essential for understanding the amount of a substance on a microscopic scale, such as in reactions involving Sb\u2082S\u2083.<\/p>\n","protected":false},"excerpt":{"rendered":" How many formula units are contained in 23.5 g of Sb2S3 The Correct Answer and Explanation is : The correct answer is: 4.17 \\times 10^{22} To determine the number of formula units in 23.5 g of antimony(III) sulfide (Sb\u2082S\u2083), follow these steps: Step 1: Calculate the molar mass of Sb\u2082S\u2083 First, we need to calculate […]<\/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-149789","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/149789","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=149789"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/149789\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=149789"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=149789"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=149789"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Calculate the molar mass of Sb\u2082S\u2083<\/h3>\n\n\n\n
\n
\n
Step 2: Calculate the number of moles of Sb\u2082S\u2083 in 23.5 g<\/h3>\n\n\n\n
\\text{moles} = \\frac{\\text{mass (g)}}{\\text{molar mass (g\/mol)}}
]<\/p>\n\n\n\n
\\text{moles of Sb\u2082S\u2083} = \\frac{23.5\\ \\text{g}}{339.70\\ \\text{g\/mol}} \\approx 0.0692\\ \\text{moles}
]<\/p>\n\n\n\nStep 3: Use Avogadro’s number to find the number of formula units<\/h3>\n\n\n\n
\\text{formula units} = 0.0692\\ \\text{moles} \\times 6.022 \\times 10^{23}\\ \\text{units\/mole}
]<\/p>\n\n\n\n
\\text{formula units} \\approx 4.17 \\times 10^{22}
]<\/p>\n\n\n\nExplanation (around 300 words)<\/h3>\n\n\n\n