How many moles Cl ion were present in the sample? (Use Eqs. 2 and 3.)<\/p>\n\n\n\n
a. moles Cl
b. How many grams Cl ion were present? (Use Eq. 4.)
What was the mass percent Cl ion in the sample? (Use Eq. 5.)
c. % C<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n To answer this question thoroughly, I need to know the values in Equations 2, 3, 4, and 5<\/strong>, or details from the experiment\/sample you\u2019re referring to (e.g., volume, molarity, or titration data). However, I can walk you through a general step-by-step method<\/strong> using standard chemical principles<\/strong> and formulas based on typical usage of such equations for calculating moles, grams, and mass percent of chloride (Cl\u207b) ions.<\/p>\n\n\n\n Let\u2019s assume you\u2019re analyzing a sample for chloride ion (Cl\u207b) using Mohr titration<\/strong> with silver nitrate (AgNO\u2083), where:<\/p>\n\n\n\n Eq. 2<\/strong>: Eq. 3<\/strong>: Eq. 4<\/strong>: Eq. 5<\/strong>: Let\u2019s say:<\/p>\n\n\n\n Using Eq. 2 and 3:<\/p>\n\n\n\n $$ $$ Using Eq. 4:<\/p>\n\n\n\n $$ Using Eq. 5:<\/p>\n\n\n\n $$ In chemical analysis, determining the amount of a specific ion in a sample is crucial for understanding composition and purity. Here, we calculate the amount of chloride ion (Cl\u207b)<\/strong> in a given sample using a titration approach with silver nitrate (AgNO\u2083), a classic method known as Mohr titration<\/strong>. Silver nitrate reacts with chloride ions to form an insoluble precipitate of silver chloride (AgCl), and this stoichiometric 1:1 reaction forms the basis of the calculation.<\/p>\n\n\n\n First, we determine the number of moles of silver nitrate added during titration using molarity (mol\/L)<\/strong> multiplied by the volume (L)<\/strong>, as described in Equation 2. Since each mole of AgNO\u2083 reacts with exactly one mole of Cl\u207b, the moles of AgNO\u2083 used equals the moles of Cl\u207b in the sample (Equation 3).<\/p>\n\n\n\n Next, the moles of Cl\u207b are converted to grams using the molar mass of Cl\u207b (35.45 g\/mol), a standard atomic value from the periodic table (Equation 4). This gives the actual mass of chloride ion present in the sample.<\/p>\n\n\n\n Lastly, to determine the mass percent<\/strong>, we divide the mass of Cl\u207b by the total sample mass and multiply by 100 (Equation 5). This value reflects the proportion of chloride ion in the entire sample<\/strong>, a critical parameter in quality control and chemical formulations.<\/p>\n\n\n\n These steps illustrate the essential role of stoichiometry and unit conversions in analytical chemistry. Whether in water quality testing, food analysis, or industrial processes, knowing how to compute the mass percent of ions like Cl\u207b ensures accurate labeling and regulatory compliance.<\/p>\n\n\n\n If you provide the actual values or equations used in your specific scenario, I can tailor the answer further.<\/p>\n","protected":false},"excerpt":{"rendered":" How many moles Cl ion were present in the sample? (Use Eqs. 2 and 3.) a. moles Clb. How many grams Cl ion were present? (Use Eq. 4.)What was the mass percent Cl ion in the sample? (Use Eq. 5.)c. % C The Correct Answer and Explanation is: To answer this question thoroughly, I need […]<\/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-217294","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217294","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=217294"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217294\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=217294"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=217294"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=217294"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nAssumptions and General Formulas<\/strong><\/h3>\n\n\n\n
$\\text{mol AgNO}_3 = M \\times V$
This gives the moles of AgNO\u2083 used.<\/p>\n\n\n\n
$\\text{mol Cl}^- = \\text{mol AgNO}_3$
Since Ag\u207a reacts with Cl\u207b in a 1:1 molar ratio:
$\\text{Ag}^+ + \\text{Cl}^- \\rightarrow \\text{AgCl} \\downarrow$<\/p>\n\n\n\n
$\\text{mass of Cl}^- = \\text{mol Cl}^- \\times \\text{molar mass of Cl}^-$
Molar mass of Cl\u207b = 35.45 g\/mol.<\/p>\n\n\n\n
$\\text{Mass percent Cl}^- = \\left( \\frac{\\text{mass of Cl}^-}{\\text{mass of sample}} \\right) \\times 100$<\/p>\n\n\n\n
\n\n\n\nSample Calculation<\/strong><\/h3>\n\n\n\n
\n
a. Moles of Cl\u207b<\/h4>\n\n\n\n
\\text{mol AgNO}_3 = 0.100 \\, \\text{mol\/L} \\times 0.02500 \\, \\text{L} = 0.002500 \\, \\text{mol}
$$<\/p>\n\n\n\n
\\text{mol Cl}^- = \\text{mol AgNO}_3 = 0.002500 \\, \\text{mol}
$$<\/p>\n\n\n\nb. Mass of Cl\u207b<\/h4>\n\n\n\n
\\text{mass Cl}^- = 0.002500 \\, \\text{mol} \\times 35.45 \\, \\text{g\/mol} = 0.08863 \\, \\text{g}
$$<\/p>\n\n\n\nc. Mass Percent of Cl\u207b<\/h4>\n\n\n\n
\\text{mass percent Cl}^- = \\left( \\frac{0.08863}{0.5000} \\right) \\times 100 = 17.73\\%
$$<\/p>\n\n\n\n
\n\n\n\nExplanation (300+ words)<\/strong><\/h3>\n\n\n\n