To answer this question, we need to refer to Table 9.1<\/strong>, which typically lists the solubility of various salts (like KCl) in water at different temperatures. A commonly accepted version of this table shows the solubility of KCl (potassium chloride)<\/strong> as follows:<\/p>\n\n\n\n From the table: Solution mass = mass of solute + mass of water Mass percent=(mass of solutemass of solution)\u00d7100=(42.6142.6)\u00d7100\u224829.87%\\text{Mass percent} = \\left( \\frac{\\text{mass of solute}}{\\text{mass of solution}} \\right) \\times 100 = \\left( \\frac{42.6}{142.6} \\right) \\times 100 \\approx 29.87\\%Mass percent=(mass of solutionmass of solute\u200b)\u00d7100=(142.642.6\u200b)\u00d7100\u224829.87%<\/p>\n\n\n\n At 100\u00b0C, solubility = 54.0 g per 100 g water At 100\u00b0C: 54.0 g dissolved This problem centers around understanding the temperature-dependent solubility<\/strong> of potassium chloride (KCl) in water. Solubility increases with temperature, a key property of most solid solutes. At 50\u00b0C, 42.6 grams<\/strong> of KCl dissolve in 100 grams of water, indicating this is the saturation point\u2014no more KCl can dissolve at this temperature without changing the amount of water or the temperature.<\/p>\n\n\n\n The total mass of the solution is the sum of the solute and solvent: 142.6 g<\/strong>. To determine concentration, we calculate the mass percent<\/strong>, a common expression of solution composition: the solute\u2019s mass divided by the total solution mass, multiplied by 100. This gives about 29.87%<\/strong>, meaning nearly one-third of the solution\u2019s mass is KCl.<\/p>\n\n\n\n If the saturated solution at 50\u00b0C is heated to 100\u00b0C, it can now dissolve up to 54.0 g<\/strong> of KCl. Since 42.6 g are already dissolved, an additional 11.4 g<\/strong> can dissolve. This is useful in industrial processes that rely on temperature shifts to change solubility.<\/p>\n\n\n\n Finally, if the solution saturated at 100\u00b0C is cooled to 30\u00b0C, where solubility is only 36.0 g, 18.0 g of KCl will precipitate out<\/strong>. This reversible behavior\u2014dissolution and precipitation due to temperature change\u2014is fundamental in crystallization techniques<\/strong>, such as purifying compounds in chemical labs or salt harvesting in industry.<\/p>\n\n\n\n Solution of KCI is saturated at 50\u00c2\u00b0C. Use Table 9.1. How many grams of solute are dissolved in 100 g of water? (6) What is the total mass of the solution? What is the mass percent of this solution at 50\u00c2\u00b0C? (d) If the solution is heated to 100\u00c2\u00b0C, how much more KCI can be […]<\/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-228304","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/228304","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=228304"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/228304\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=228304"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=228304"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=228304"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Temperature (\u00b0C)<\/th> Solubility of KCl (g per 100 g H\u2082O)<\/th><\/tr><\/thead> 30<\/td> 36.0<\/td><\/tr> 50<\/td> 42.6<\/td><\/tr> 100<\/td> 54.0<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n (a) How many grams of solute are dissolved in 100 g of water at 50\u00b0C?<\/strong><\/h3>\n\n\n\n
42.6 g<\/strong> of KCl dissolve in 100 g<\/strong> of water at 50\u00b0C<\/strong>.<\/p>\n\n\n\n
\n\n\n\n(b) What is the total mass of the solution?<\/strong><\/h3>\n\n\n\n
= 42.6 g + 100 g
= 142.6 g<\/strong><\/p>\n\n\n\n
\n\n\n\n(c) What is the mass percent of this solution at 50\u00b0C?<\/strong><\/h3>\n\n\n\n
\n\n\n\n(d) If the solution is heated to 100\u00b0C, how much more KCl can be dissolved in the solution without adding more water?<\/strong><\/h3>\n\n\n\n
Already dissolved at 50\u00b0C = 42.6 g
Additional KCl that can be dissolved = 54.0 \u2212 42.6 = 11.4 g<\/strong><\/p>\n\n\n\n
\n\n\n\n(e) If saturated at 100\u00b0C and cooled to 30\u00b0C, how much KCl will precipitate out?<\/strong><\/h3>\n\n\n\n
At 30\u00b0C: 36.0 g can remain dissolved
Amount that precipitates = 54.0 \u2212 36.0 = 18.0 g<\/strong><\/p>\n\n\n\n
\n\n\n\n\u2705 Summary Explanation :<\/h3>\n\n\n\n
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