A sample of sodium sulfite (Na2SO3) has a mass of 2.25 g.
a. How many Na+ ions are present?
b. How many SO3\u00b2? ions are present?
c. What is the mass in grams of one formula unit of Na2SO3?<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n Let’s solve each part of the question step by step.<\/p>\n\n\n\n Step 1: Molar mass of Na\u2082SO\u2083<\/strong> The formula for sodium sulfite is Na\u2082SO\u2083, which contains:<\/p>\n\n\n\n So, the molar mass of Na\u2082SO\u2083 is:<\/p>\n\n\n\n [ a. How many Na\u207a ions are present?<\/strong><\/p>\n\n\n\n [ [ b. How many SO\u2083\u00b2\u207b ions are present?<\/strong><\/p>\n\n\n\n [ c. What is the mass in grams of one formula unit of Na\u2082SO\u2083?<\/strong><\/p>\n\n\n\n To find the mass of one formula unit of Na\u2082SO\u2083, we need to use the molar mass. The molar mass of Na\u2082SO\u2083 is 126.05 g\/mol, and this corresponds to 1 mole of Na\u2082SO\u2083, which contains Avogadro\u2019s number of formula units.<\/p>\n\n\n\n The mass of one formula unit is:<\/p>\n\n\n\n [ This problem involved converting the mass of Na\u2082SO\u2083 to moles, then using stoichiometry to find the number of ions, and finally determining the mass of a single formula unit based on the molar mass and Avogadro\u2019s number.<\/p>\n","protected":false},"excerpt":{"rendered":" A sample of sodium sulfite (Na2SO3) has a mass of 2.25 g.a. How many Na+ ions are present?b. How many SO3\u00b2? ions are present?c. What is the mass in grams of one formula unit of Na2SO3? The correct answer and explanation is : Let’s solve each part of the question step by step. Given: Step […]<\/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-204655","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/204655","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=204655"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/204655\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=204655"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=204655"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=204655"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Given:<\/h3>\n\n\n\n
\n
First, we need the molar mass of Na\u2082SO\u2083. To calculate this, we add the atomic masses of the elements involved:<\/p>\n\n\n\n\n
\n
\\text{Molar mass of Na\u2082SO\u2083} = (2 \\times 22.99) + (1 \\times 32.07) + (3 \\times 16.00)
]
[
\\text{Molar mass of Na\u2082SO\u2083} = 45.98 + 32.07 + 48.00 = 126.05 \\, \\text{g\/mol}
]<\/p>\n\n\n\n\n
\\text{Moles of Na\u2082SO\u2083} = \\frac{\\text{mass of Na\u2082SO\u2083}}{\\text{molar mass of Na\u2082SO\u2083}} = \\frac{2.25 \\, \\text{g}}{126.05 \\, \\text{g\/mol}} = 0.0178 \\, \\text{mol}
]<\/p>\n\n\n\n\n
\\text{Number of Na\u207a ions} = 2 \\times \\text{moles of Na\u2082SO\u2083} \\times \\text{Avogadro’s number}
]
[
\\text{Number of Na\u207a ions} = 2 \\times 0.0178 \\, \\text{mol} \\times 6.022 \\times 10^{23} \\, \\text{ions\/mol}
]
[
\\text{Number of Na\u207a ions} = 2.14 \\times 10^{22} \\, \\text{Na\u207a ions}
]<\/p>\n\n\n\n\n
\\text{Number of SO\u2083\u00b2\u207b ions} = \\text{moles of Na\u2082SO\u2083} \\times \\text{Avogadro’s number}
]
[
\\text{Number of SO\u2083\u00b2\u207b ions} = 0.0178 \\, \\text{mol} \\times 6.022 \\times 10^{23} \\, \\text{ions\/mol}
]
[
\\text{Number of SO\u2083\u00b2\u207b ions} = 1.07 \\times 10^{22} \\, \\text{SO\u2083\u00b2\u207b ions}
]<\/p>\n\n\n\n
\\text{Mass of one formula unit} = \\frac{\\text{Molar mass of Na\u2082SO\u2083}}{\\text{Avogadro’s number}} = \\frac{126.05 \\, \\text{g\/mol}}{6.022 \\times 10^{23} \\, \\text{formula units\/mol}}
]
[
\\text{Mass of one formula unit} = 2.09 \\times 10^{-22} \\, \\text{g}
]<\/p>\n\n\n\nSummary of answers:<\/h3>\n\n\n\n
\n