{"id":261367,"date":"2025-07-20T05:47:10","date_gmt":"2025-07-20T05:47:10","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=261367"},"modified":"2025-07-20T05:47:12","modified_gmt":"2025-07-20T05:47:12","slug":"the-molar-heat-of-solution-ihsoln-of-sodium-hydroxide-is-445-1-kj-mol-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/20\/the-molar-heat-of-solution-ihsoln-of-sodium-hydroxide-is-445-1-kj-mol-2\/","title":{"rendered":"The molar heat of solution (\u00ce\u201dHsoln) of sodium hydroxide is -445.1 kJ\/mol."},"content":{"rendered":"\n

The molar heat of solution (\u00ce\u201dHsoln) of sodium hydroxide is -445.1 kJ\/mol. In a certain experiment, 5.00 g of solid sodium hydroxide is completely dissolved in 1.000 L of 20.0 \u00c2\u00b0C water in a foam cup calorimeter. Assuming no heat loss, calculate the final temperature of the water.<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To calculate the final temperature of the water when sodium hydroxide (NaOH) dissolves, we can use the heat released by the dissolution process. The heat released will raise the temperature of the water. Here\u2019s the step-by-step solution:<\/p>\n\n\n\n

Step 1: Calculate the moles of NaOH<\/h3>\n\n\n\n

The given mass of NaOH is 5.00 g, and its molar mass is approximately 40.0 g\/mol. To find the number of moles of NaOH: moles of NaOH=mass of NaOHmolar mass of NaOH=5.00\u2009g40.0\u2009g\/mol=0.125\u2009mol\\text{moles of NaOH} = \\frac{\\text{mass of NaOH}}{\\text{molar mass of NaOH}} = \\frac{5.00 \\, \\text{g}}{40.0 \\, \\text{g\/mol}} = 0.125 \\, \\text{mol}moles of NaOH=molar mass of NaOHmass of NaOH\u200b=40.0g\/mol5.00g\u200b=0.125mol<\/p>\n\n\n\n

Step 2: Calculate the heat released during the dissolution<\/h3>\n\n\n\n

The heat of solution (\u0394Hsoln) for NaOH is -445.1 kJ\/mol. This value tells us how much heat is released per mole of NaOH dissolved. Since NaOH dissociates in water, the heat released is: heat released=moles of NaOH\u00d7\u0394Hsoln=0.125\u2009mol\u00d7(\u2212445.1\u2009kJ\/mol)=\u221255.64\u2009kJ\\text{heat released} = \\text{moles of NaOH} \\times \\Delta H_{\\text{soln}} = 0.125 \\, \\text{mol} \\times (-445.1 \\, \\text{kJ\/mol}) = -55.64 \\, \\text{kJ}heat released=moles of NaOH\u00d7\u0394Hsoln\u200b=0.125mol\u00d7(\u2212445.1kJ\/mol)=\u221255.64kJ<\/p>\n\n\n\n

Thus, 55.64 kJ of heat is released to the water.<\/p>\n\n\n\n

Step 3: Calculate the temperature change of the water<\/h3>\n\n\n\n

The heat released will increase the temperature of the water. The specific heat capacity of water is 4.18\u2009J\/g\u00b0C4.18 \\, \\text{J\/g\u00b0C}4.18J\/g\u00b0C, and the mass of the water is 1.000 L, which is equivalent to 1000 g of water. We can now use the heat equation: q=m\u22c5c\u22c5\u0394Tq = m \\cdot c \\cdot \\Delta Tq=m\u22c5c\u22c5\u0394T<\/p>\n\n\n\n

Where:<\/p>\n\n\n\n

    \n
  • qqq is the heat absorbed by the water, 55.64 kJ = 55640 J,<\/li>\n\n\n\n
  • mmm is the mass of the water, 1000 g,<\/li>\n\n\n\n
  • ccc is the specific heat capacity of water, 4.18\u2009J\/g\u00b0C4.18 \\, \\text{J\/g\u00b0C}4.18J\/g\u00b0C,<\/li>\n\n\n\n
  • \u0394T\\Delta T\u0394T is the change in temperature.<\/li>\n<\/ul>\n\n\n\n

    Rearranging the equation to solve for \u0394T\\Delta T\u0394T: \u0394T=qm\u22c5c=55640\u2009J1000\u2009g\u00d74.18\u2009J\/g\u00b0C=13.3\u00b0C\\Delta T = \\frac{q}{m \\cdot c} = \\frac{55640 \\, \\text{J}}{1000 \\, \\text{g} \\times 4.18 \\, \\text{J\/g\u00b0C}} = 13.3\u00b0C\u0394T=m\u22c5cq\u200b=1000g\u00d74.18J\/g\u00b0C55640J\u200b=13.3\u00b0C<\/p>\n\n\n\n

    Step 4: Calculate the final temperature<\/h3>\n\n\n\n

    The initial temperature of the water is 20.0\u00b0C. Adding the temperature change: final temperature=20.0\u00b0C+13.3\u00b0C=33.3\u00b0C\\text{final temperature} = 20.0\u00b0C + 13.3\u00b0C = 33.3\u00b0Cfinal temperature=20.0\u00b0C+13.3\u00b0C=33.3\u00b0C<\/p>\n\n\n\n

    Final Answer:<\/h3>\n\n\n\n

    The final temperature of the water will be 33.3\u00b0C<\/strong>.<\/p>\n\n\n\n

    This calculation assumes no heat loss to the surroundings and that all the heat released by the dissolving NaOH is absorbed by the water.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    The molar heat of solution (\u00ce\u201dHsoln) of sodium hydroxide is -445.1 kJ\/mol. In a certain experiment, 5.00 g of solid sodium hydroxide is completely dissolved in 1.000 L of 20.0 \u00c2\u00b0C water in a foam cup calorimeter. Assuming no heat loss, calculate the final temperature of the water. The Correct Answer and Explanation is: To […]<\/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-261367","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/261367","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=261367"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/261367\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=261367"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=261367"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=261367"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}