{"id":29493,"date":"2025-06-21T05:10:44","date_gmt":"2025-06-21T05:10:44","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=29493"},"modified":"2025-06-21T05:10:46","modified_gmt":"2025-06-21T05:10:46","slug":"assuming-complete-dissociation-calculate-the-freezing-point-of-a-solution-prepared-by-dissolving-6g-of-na2so4-a%c2%b7","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/assuming-complete-dissociation-calculate-the-freezing-point-of-a-solution-prepared-by-dissolving-6g-of-na2so4-a%c2%b7\/","title":{"rendered":"Assuming complete dissociation, calculate the freezing point of a solution prepared by dissolving 6g of Na2SO4 \u00c2\u00b7"},"content":{"rendered":"\n<p>Assuming complete dissociation, calculate the freezing point of a solution prepared by dissolving 6g of Na2SO4 \u00c2\u00b7 10H2O.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To calculate the <strong>freezing point<\/strong> of a solution prepared by dissolving 6 g of sodium sulfate decahydrate (Na\u2082SO\u2084\u00b710H\u2082O), we use the <strong>freezing point depression formula<\/strong>:\u0394Tf=i\u22c5Kf\u22c5m\\Delta T_f = i \\cdot K_f \\cdot m\u0394Tf\u200b=i\u22c5Kf\u200b\u22c5m<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u0394Tf\\Delta T_f\u0394Tf\u200b is the freezing point depression<\/li>\n\n\n\n<li>iii is the van\u2019t Hoff factor (number of particles the solute dissociates into)<\/li>\n\n\n\n<li>KfK_fKf\u200b is the freezing point depression constant of water (1.86 \u00b0C\u00b7kg\/mol)<\/li>\n\n\n\n<li>mmm is the molality of the solution<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate molar mass of Na\u2082SO\u2084\u00b710H\u2082O<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Na = 22.99 \u00d7 2 = 45.98 g\/mol<\/li>\n\n\n\n<li>S = 32.07 g\/mol<\/li>\n\n\n\n<li>O\u2084 = 16.00 \u00d7 4 = 64.00 g\/mol<\/li>\n\n\n\n<li>10H\u2082O = 10 \u00d7 (2\u00d71.008 + 16.00) = 10 \u00d7 18.016 = 180.16 g\/mol<\/li>\n<\/ul>\n\n\n\n<p><strong>Total molar mass<\/strong> = 45.98 + 32.07 + 64.00 + 180.16 = <strong>322.21 g\/mol<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Find moles of solute<\/h3>\n\n\n\n<p>Moles&nbsp;of&nbsp;Na\u2082SO\u2084\\cdotp10H\u2082O=6&nbsp;g322.21&nbsp;g\/mol=0.01862&nbsp;mol\\text{Moles of Na\u2082SO\u2084\u00b710H\u2082O} = \\frac{6\\text{ g}}{322.21\\text{ g\/mol}} = 0.01862 \\text{ mol}Moles&nbsp;of&nbsp;Na\u2082SO\u2084\\cdotp10H\u2082O=322.21&nbsp;g\/mol6&nbsp;g\u200b=0.01862&nbsp;mol<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Assume solution is made in 100 g (0.1 kg) of water<\/h3>\n\n\n\n<p>Molality m=0.018620.1=0.1862\u2009mol\/kgm = \\frac{0.01862}{0.1} = 0.1862 \\, \\text{mol\/kg}m=0.10.01862\u200b=0.1862mol\/kg<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Van\u2019t Hoff factor iii<\/h3>\n\n\n\n<p>Na\u2082SO\u2084 dissociates completely into 2 Na\u207a and 1 SO\u2084\u00b2\u207bi=3i = 3i=3<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 5: Calculate \u0394Tf\\Delta T_f\u0394Tf\u200b<\/h3>\n\n\n\n<p>\\Delta T_f = 3 \\cdot 1.86 \\cdot 0.1862 = 1.039\\,^\\circ \\text{C}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 6: Freezing point of water = 0 \u00b0C<\/h3>\n\n\n\n<p>Freezing point of solution =0\u22121.039=\u22121.04\u2218C0 &#8211; 1.039 = \\boxed{-1.04^\\circ \\text{C}}0\u22121.039=\u22121.04\u2218C\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation <\/h3>\n\n\n\n<p>The freezing point depression is a colligative property, meaning it depends on the number of solute particles in a solution and not their identity. In this problem, we\u2019re dissolving 6 grams of sodium sulfate decahydrate (Na\u2082SO\u2084\u00b710H\u2082O) in water. This hydrated salt fully dissociates in water into three ions per formula unit: two sodium ions (Na\u207a) and one sulfate ion (SO\u2084\u00b2\u207b), giving it a van\u2019t Hoff factor of 3. This factor is essential in determining how much the freezing point will be lowered.<\/p>\n\n\n\n<p>First, we calculate the molar mass of the hydrate. Each molecule includes the base salt Na\u2082SO\u2084 plus ten water molecules. The full molar mass is approximately 322.21 g\/mol. Using this, we determine the number of moles of the substance in the solution by dividing 6 grams by the molar mass, giving about 0.01862 moles. Assuming the salt is dissolved in 100 grams of water (a common assumption unless otherwise stated), we calculate molality by dividing the moles of solute by kilograms of solvent. This gives a molality of 0.1862 mol\/kg.<\/p>\n\n\n\n<p>Next, applying the freezing point depression formula, we multiply the van\u2019t Hoff factor (3), the molality (0.1862 mol\/kg), and the cryoscopic constant of water (1.86 \u00b0C\u00b7kg\/mol). This yields a freezing point depression of 1.039 \u00b0C. Finally, subtracting this from the normal freezing point of water (0 \u00b0C) gives a final freezing point of approximately -1.04 \u00b0C.<\/p>\n\n\n\n<p>This shows how ionic compounds can significantly depress the freezing point due to the number of particles formed in solution.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-312.jpeg\" alt=\"\" class=\"wp-image-29494\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-312.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-312-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-312-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Assuming complete dissociation, calculate the freezing point of a solution prepared by dissolving 6g of Na2SO4 \u00c2\u00b7 10H2O. The Correct Answer and Explanation is: To calculate the freezing point of a solution prepared by dissolving 6 g of sodium sulfate decahydrate (Na\u2082SO\u2084\u00b710H\u2082O), we use the freezing point depression formula:\u0394Tf=i\u22c5Kf\u22c5m\\Delta T_f = i \\cdot K_f \\cdot [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-29493","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/29493","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=29493"}],"version-history":[{"count":2,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/29493\/revisions"}],"predecessor-version":[{"id":29496,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/29493\/revisions\/29496"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=29493"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=29493"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=29493"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}