{"id":196161,"date":"2025-03-03T05:19:05","date_gmt":"2025-03-03T05:19:05","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=196161"},"modified":"2025-03-03T05:19:08","modified_gmt":"2025-03-03T05:19:08","slug":"hydrogen-bromide-hbr-is-a-strong-electrolyte-in-water","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/03\/hydrogen-bromide-hbr-is-a-strong-electrolyte-in-water\/","title":{"rendered":"Hydrogen bromide, HBr, is a strong electrolyte in water"},"content":{"rendered":"\n<p>Hydrogen bromide, HBr, is a strong electrolyte in water. What is the boiling point elevation ATb) of a solution containing 125 g of HBr in 1.75 kg of water? The molar mass of HBr is 80.9 g\/mol, and the boiling point elevation constant (K6) of water is 0.512\u00c3\u201a\u00c2\u00b0C\/m. <\/p>\n\n\n\n<p>1. molality of the solution <\/p>\n\n\n\n<p>2. boiling point elevation of the solution <\/p>\n\n\n\n<p>Give your answers in three significant figures None<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>When hydrogen bromide (HBr) dissolves in water, it dissociates completely into H\u207a and Br\u207b ions, making it a strong electrolyte. This dissociation influences the solution&#8217;s colligative properties, such as boiling point elevation.\ue206<\/p>\n\n\n\n<p><strong>1. Molality of the Solution:<\/strong><\/p>\n\n\n\n<p>Molality (m) is defined as the number of moles of solute per kilogram of solvent. To calculate the molality of the HBr solution:\ue206<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Moles of HBr:<\/strong> Given 125 g of HBr and its molar mass of 80.9 g\/mol:\ue206 [<br>\\text{Moles of HBr} = \\frac{125\\, \\text{g}}{80.9\\, \\text{g\/mol}} \\approx 1.544\\, \\text{mol}<br>]\ue206<\/li>\n\n\n\n<li><strong>Mass of Water:<\/strong> The solvent is water with a mass of 1.75 kg.\ue206<\/li>\n\n\n\n<li><strong>Molality (m):<\/strong> Calculated as moles of solute per kilogram of solvent:\ue206 [<br>m = \\frac{1.544\\, \\text{mol}}{1.75\\, \\text{kg}} \\approx 0.883\\, \\text{mol\/kg}<br>]\ue206<\/li>\n<\/ul>\n\n\n\n<p><strong>2. Boiling Point Elevation (\u0394Tb):<\/strong><\/p>\n\n\n\n<p>The boiling point elevation is given by the formula:\ue206<\/p>\n\n\n\n<p>[<br>\\Delta T_b = i \\times K_b \\times m<br>]\ue206<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>i is the van &#8216;t Hoff factor, representing the number of particles the solute dissociates into. For HBr, i = 2 (H\u207a and Br\u207b).\ue206<\/li>\n\n\n\n<li>K_b is the ebullioscopic constant of the solvent. For water, K_b = 0.512\u202f\u00b0C\u00b7kg\/mol.\ue206<\/li>\n\n\n\n<li>m is the molality of the solution.\ue206<\/li>\n<\/ul>\n\n\n\n<p>Calculating \u0394Tb:<\/p>\n\n\n\n<p>[<br>\\Delta T_b = 2 \\times 0.512\\, \\text{\u00b0C\u00b7kg\/mol} \\times 0.883\\, \\text{mol\/kg} \\approx 0.905\\, \\text{\u00b0C}<br>]\ue206<\/p>\n\n\n\n<p>Therefore, the boiling point of the solution increases by approximately 0.905\u202f\u00b0C above the boiling point of pure water. Since the boiling point of pure water is 100\u202f\u00b0C, the boiling point of this solution would be approximately 100.91\u202f\u00b0C.\ue206<\/p>\n\n\n\n<p><strong>Explanation:<\/strong><\/p>\n\n\n\n<p>Boiling point elevation is a colligative property, meaning it depends on the number of solute particles in the solution, not their identity. When a non-volatile solute like HBr is added to a solvent like water, it dissociates into ions, increasing the number of particles in the solution. This disruption of the solvent&#8217;s vapor pressure requires a higher temperature to reach the boiling point, resulting in boiling point elevation. \ue200cite\ue202turn0search2\ue201\ue206<\/p>\n\n\n\n<p>In this scenario, the complete dissociation of HBr into H\u207a and Br\u207b ions contributes to the observed increase in boiling point, demonstrating the effect of solute particle concentration on the colligative properties of solutions.\ue206<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Hydrogen bromide, HBr, is a strong electrolyte in water. What is the boiling point elevation ATb) of a solution containing 125 g of HBr in 1.75 kg of water? The molar mass of HBr is 80.9 g\/mol, and the boiling point elevation constant (K6) of water is 0.512\u00c3\u201a\u00c2\u00b0C\/m. 1. molality of the solution 2. boiling [&hellip;]<\/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-196161","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196161","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=196161"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196161\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=196161"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=196161"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=196161"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}