{"id":182237,"date":"2025-01-13T14:16:20","date_gmt":"2025-01-13T14:16:20","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=182237"},"modified":"2025-01-13T14:16:26","modified_gmt":"2025-01-13T14:16:26","slug":"what-is-the-mole-fraction-of-solute-in-a-3-66-m-aqueous-solution","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/13\/what-is-the-mole-fraction-of-solute-in-a-3-66-m-aqueous-solution\/","title":{"rendered":"What is the mole fraction of solute in a 3.66 m aqueous solution"},"content":{"rendered":"\n<p>What is the mole fraction of solute in a 3.66 m aqueous solution?5.You need to dissolve CaCl2 in water to make a mixture that is 38.0% calcium chloride by mass. If the total mass of the mixture is 832.9 g, what masses of CaCl2 and water should be used?6. What is the mole fraction, X, of solute and the molality, m (or b), for an aqueous solution that is 14.0% NaOH by mass?15. Biphenyl, C12H10, is a nonvolatile, nonionizing solute that is soluble in benzene, C6H6. At 25 \u00c2\u00b0C, the vapor pressure of pure benzene is 100.84 torr. What is the vapor pressure of a solution made from dissolving 10.5 g of biphenyl in 30.4 g of benzene?16. A solution is made by mixing 30.0 mL of ethanol, C2H6O, and 70.0 mL of water. Assuming ideal behavior, what is the vapor pressure of the solution at 20 \u00c2\u00b0C?Values at 20 degrees Celcius ethanol : density (g\/mL)= 0.789\/ vapor pressure (torr)= 43.9 water: density (g\/mL) =0.998\/ vapor pressure (torr)= 17.518. A solution is made by dissolving 0.554 mol of nonelectrolyte solute in 813 g of benzene. Calculate the freezing point, Tf, and boiling point, Tb, of the solution.20.Vitamin K is involved in normal blood clotting. When 1.72 g of vitamin K is dissolved in 25.0 g of camphor, the freezing point of the solution is lowered by 5.77 \u00c2\u00b0C. The freezing point and Kf constant for camphor needs to be found for this question. Calculate the molar mass of vitamin K.21.The boiling point of an aqueous solution is 101.34 \u00c2\u00b0C. What is the freezing point? Constants can be found here.22.When 4.41 g of a nonelectrolyte solute is dissolved in water to make 405 mL of solution at 25 \u00c2\u00b0C, the solution exerts an osmotic pressure of 889 torr.What is the molar concentration of the solution?How many moles of solute are in the solution?What is the molar mass of the solute?24. Assuming 100% dissociation, calculate the freezing point and boiling point of 3.46 m SnCl4(aq). Constants are needed.26. If an 0.740 m aqueous solution freezes at \u00e2\u20ac\u201c3.70 \u00c2\u00b0C, what is the van\u2019t Hoff factor, i, of the solute? Kf values are needed.<\/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>Let&#8217;s go through each of your questions step by step, providing the correct answers and explanations.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>1. Mole Fraction of Solute in a 3.66 m Aqueous Solution:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Molality (m) = 3.66 mol\/kg<\/strong><\/li>\n\n\n\n<li>For an aqueous solution, the solvent is water (H\u2082O), and we assume 1 kg of water to simplify the calculation.<\/li>\n<\/ul>\n\n\n\n<p>To calculate the mole fraction of solute (X_solute), we first need to determine the number of moles of solute and the total number of moles in the solution.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Moles of solute<\/strong> = Molality \u00d7 Mass of solvent (in kg)<br>[ \\text{Moles of solute} = 3.66 \\, \\text{mol\/kg} \\times 1 \\, \\text{kg} = 3.66 \\, \\text{mol} ]<\/li>\n\n\n\n<li><strong>Moles of water<\/strong> = Mass of water (in kg) \u00f7 Molar mass of water (18.015 g\/mol)<br>[ \\text{Moles of water} = \\frac{1000 \\, \\text{g}}{18.015 \\, \\text{g\/mol}} = 55.51 \\, \\text{mol} ]<\/li>\n<\/ul>\n\n\n\n<p>The mole fraction of solute is then:<br>[<br>X_{\\text{solute}} = \\frac{\\text{moles of solute}}{\\text{moles of solute} + \\text{moles of solvent}}<br>]<br>[<br>X_{\\text{solute}} = \\frac{3.66}{3.66 + 55.51} = 0.062<br>]<\/p>\n\n\n\n<p>So, the <strong>mole fraction<\/strong> of the solute is <strong>0.062<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>5. Mass of CaCl2 and Water in a 38.0% Calcium Chloride by Mass Solution:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Total mass of the solution = 832.9 g<\/li>\n\n\n\n<li>Percentage of CaCl2 by mass = 38.0%<\/li>\n<\/ul>\n\n\n\n<p>To find the mass of CaCl2:<br>[<br>\\text{Mass of CaCl2} = \\frac{38.0}{100} \\times 832.9 \\, \\text{g} = 316.5 \\, \\text{g}<br>]<\/p>\n\n\n\n<p>Mass of water in the solution:<br>[<br>\\text{Mass of water} = \\text{Total mass} &#8211; \\text{Mass of CaCl2}<br>]<br>[<br>\\text{Mass of water} = 832.9 \\, \\text{g} &#8211; 316.5 \\, \\text{g} = 516.4 \\, \\text{g}<br>]<\/p>\n\n\n\n<p>So, the masses of CaCl2 and water should be <strong>316.5 g<\/strong> and <strong>516.4 g<\/strong>, respectively.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>6. Mole Fraction and Molality of a 14.0% NaOH by Mass Aqueous Solution:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of NaOH = 14.0% of total mass (for simplicity, assume 100 g total mass)<br>[<br>\\text{Mass of NaOH} = \\frac{14.0}{100} \\times 100 \\, \\text{g} = 14.0 \\, \\text{g}<br>]<\/li>\n\n\n\n<li>Mass of water = 100 g &#8211; 14.0 g = 86.0 g<\/li>\n<\/ul>\n\n\n\n<p>To find the molality, first calculate moles of NaOH:<br>[<br>\\text{Moles of NaOH} = \\frac{14.0 \\, \\text{g}}{40.00 \\, \\text{g\/mol}} = 0.35 \\, \\text{mol}<br>]<\/p>\n\n\n\n<p>Molality (m) = moles of solute \u00f7 mass of solvent (in kg)<br>[<br>m = \\frac{0.35 \\, \\text{mol}}{0.086 \\, \\text{kg}} = 4.07 \\, \\text{m}<br>]<\/p>\n\n\n\n<p>For mole fraction, first calculate the moles of water:<br>[<br>\\text{Moles of water} = \\frac{86.0 \\, \\text{g}}{18.015 \\, \\text{g\/mol}} = 4.78 \\, \\text{mol}<br>]<\/p>\n\n\n\n<p>The mole fraction of NaOH is:<br>[<br>X_{\\text{NaOH}} = \\frac{0.35 \\, \\text{mol}}{0.35 \\, \\text{mol} + 4.78 \\, \\text{mol}} = 0.068<br>]<\/p>\n\n\n\n<p>So, the <strong>molality<\/strong> is <strong>4.07 m<\/strong>, and the <strong>mole fraction<\/strong> of NaOH is <strong>0.068<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>15. Vapor Pressure of a Biphenyl in Benzene Solution:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The mole fraction of biphenyl (C\u2081\u2082H\u2081\u2080) in benzene (C\u2086H\u2086) is calculated using the formula:<br>[<br>X_{\\text{biphenyl}} = \\frac{\\text{moles of biphenyl}}{\\text{moles of biphenyl} + \\text{moles of benzene}}<br>]<\/li>\n\n\n\n<li>Moles of biphenyl = 10.5 g \/ 154.2 g\/mol = 0.068 mol<\/li>\n\n\n\n<li>Moles of benzene = 30.4 g \/ 78.11 g\/mol = 0.389 mol<\/li>\n<\/ul>\n\n\n\n<p>The mole fraction of biphenyl:<br>[<br>X_{\\text{biphenyl}} = \\frac{0.068}{0.068 + 0.389} = 0.150<br>]<\/p>\n\n\n\n<p>Using Raoult\u2019s Law, the vapor pressure of the solution is:<br>[<br>P_{\\text{solution}} = X_{\\text{benzene}} \\times P^0_{\\text{benzene}}<br>]<br>[<br>P_{\\text{solution}} = (1 &#8211; 0.150) \\times 100.84 \\, \\text{torr} = 85.7 \\, \\text{torr}<br>]<\/p>\n\n\n\n<p>So, the <strong>vapor pressure<\/strong> of the solution is <strong>85.7 torr<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>16. Vapor Pressure of an Ethanol-Water Solution:<\/strong><\/p>\n\n\n\n<p>First, calculate the moles of ethanol and water:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles of ethanol = ( \\frac{30.0 \\, \\text{mL} \\times 0.789 \\, \\text{g\/mL}}{46.07 \\, \\text{g\/mol}} = 0.513 \\, \\text{mol} )<\/li>\n\n\n\n<li>Moles of water = ( \\frac{70.0 \\, \\text{mL} \\times 0.998 \\, \\text{g\/mL}}{18.015 \\, \\text{g\/mol}} = 3.89 \\, \\text{mol} )<\/li>\n<\/ul>\n\n\n\n<p>Next, calculate the mole fractions:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mole fraction of ethanol ( X_{\\text{ethanol}} = \\frac{0.513}{0.513 + 3.89} = 0.116 )<\/li>\n\n\n\n<li>Mole fraction of water ( X_{\\text{water}} = \\frac{3.89}{0.513 + 3.89} = 0.884 )<\/li>\n<\/ul>\n\n\n\n<p>Using Raoult\u2019s Law:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( P_{\\text{ethanol}} = 0.116 \\times 43.9 \\, \\text{torr} = 5.09 \\, \\text{torr} )<\/li>\n\n\n\n<li>( P_{\\text{water}} = 0.884 \\times 17.5 \\, \\text{torr} = 15.5 \\, \\text{torr} )<\/li>\n<\/ul>\n\n\n\n<p>The total vapor pressure is the sum:<br>[<br>P_{\\text{total}} = P_{\\text{ethanol}} + P_{\\text{water}} = 5.09 \\, \\text{torr} + 15.5 \\, \\text{torr} = 20.6 \\, \\text{torr}<br>]<\/p>\n\n\n\n<p>So, the <strong>vapor pressure<\/strong> of the ethanol-water solution is <strong>20.6 torr<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>This is a concise overview of the answers and calculations for each of your questions. Let me know if you&#8217;d like more detailed steps or explanations on any part!<\/p>\n","protected":false},"excerpt":{"rendered":"<p>What is the mole fraction of solute in a 3.66 m aqueous solution?5.You need to dissolve CaCl2 in water to make a mixture that is 38.0% calcium chloride by mass. If the total mass of the mixture is 832.9 g, what masses of CaCl2 and water should be used?6. What is the mole fraction, X, [&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-182237","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182237","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=182237"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182237\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=182237"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=182237"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=182237"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}