{"id":16959,"date":"2025-06-12T06:43:02","date_gmt":"2025-06-12T06:43:02","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=16959"},"modified":"2025-06-12T06:43:10","modified_gmt":"2025-06-12T06:43:10","slug":"an-amount-of-kmno4-was-added-to-2500-ml-of-water","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/an-amount-of-kmno4-was-added-to-2500-ml-of-water\/","title":{"rendered":"An amount of KMnO4 was added to 2500 mL of water"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">An amount of KMnO4 was added to 2500 mL of water. The boiling point of the solution is now 375.25 K. What mass of KMnO4 (in grams) was added to the water? (Kb water = 0.512 \u00c2\u00b0C\/m, MW KMnO4 = 158 g\/mol, density of water = 1.0 g\/mL, boiling point of water = 373.15 K)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\"><strong>The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">To find the mass of KMnO\u2084 added, we can use the <strong>boiling point elevation<\/strong> formula: \u0394Tb=Kb\u22c5m\\Delta T_b = K_b \\cdot m<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u0394Tb\\Delta T_b is the <strong>boiling point elevation<\/strong><\/li>\n\n\n\n<li>KbK_b is the <strong>boiling point elevation constant<\/strong> for water (0.512 \u00b0C\/m)<\/li>\n\n\n\n<li>mm is the <strong>molality<\/strong> of the solution (mol solute \/ kg solvent)<\/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\"><strong>Step 1: Calculate \u0394Tb\\Delta T_b<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">\u0394Tb=375.25\u2009K\u2212373.15\u2009K=2.10\u2009\u00b0C\\Delta T_b = 375.25\\,K &#8211; 373.15\\,K = 2.10\\,\u00b0C<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 2: Calculate molality (m)<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">m=\u0394TbKb=2.100.512\u22484.10\u2009mol\/kgm = \\frac{\\Delta T_b}{K_b} = \\frac{2.10}{0.512} \\approx 4.10 \\,\\text{mol\/kg}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 3: Convert 2500 mL of water to kg<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Given the <strong>density of water is 1.0 g\/mL<\/strong>, then: 2500\u2009mL=2500\u2009g=2.500\u2009kg2500\\,\\text{mL} = 2500\\,\\text{g} = 2.500\\,\\text{kg}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 4: Calculate moles of KMnO\u2084<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">mol=m\u00d7kg&nbsp;water=4.10\u00d72.500=10.25\u2009mol\\text{mol} = m \\times \\text{kg water} = 4.10 \\times 2.500 = 10.25\\,\\text{mol}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 5: Convert moles to grams<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">mass=mol\u00d7molar&nbsp;mass=10.25\u00d7158=1619.5\u2009g\\text{mass} = \\text{mol} \\times \\text{molar mass} = 10.25 \\times 158 = \\boxed{1619.5\\,\\text{g}}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Final Answer: 1619.5&nbsp;grams&nbsp;of&nbsp;KMnO4\\boxed{1619.5 \\text{ grams of KMnO}_4}<\/strong><\/h3>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation (\u2248300 words):<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The problem involves a classic colligative property: <strong>boiling point elevation<\/strong>, which occurs when a solute is added to a solvent. In this case, potassium permanganate (KMnO\u2084) is added to water, increasing its boiling point.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">To solve this, we start by calculating how much the boiling point has changed. The normal boiling point of water is 373.15 K, and the new boiling point is 375.25 K. This gives a boiling point elevation (\u0394Tb\\Delta T_b) of 2.10 \u00b0C.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Next, we use the boiling point elevation formula, \u0394Tb=Kb\u22c5m\\Delta T_b = K_b \\cdot m, where KbK_b is a known constant for water (0.512 \u00b0C\/m). Rearranging the formula allows us to solve for molality, the number of moles of solute per kilogram of solvent. Dividing 2.10 \u00b0C by 0.512 \u00b0C\/m gives us a molality of 4.10 mol\/kg.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Since the volume of water used is 2500 mL, and the density of water is 1.0 g\/mL, the mass of the water is 2500 g or 2.500 kg. Multiplying the molality by the mass of water in kg gives us the number of moles of KMnO\u2084: 4.10 mol\/kg \u00d7 2.500 kg = 10.25 mol.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Finally, we convert moles to grams using the molar mass of KMnO\u2084 (158 g\/mol). Multiplying gives 10.25 mol \u00d7 158 g\/mol = <strong>1619.5 grams<\/strong>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This calculation assumes ideal solution behavior and that KMnO\u2084 does not dissociate significantly in a way that affects colligative properties (which is reasonable for this level of problem).<\/p>\n","protected":false},"excerpt":{"rendered":"<p>An amount of KMnO4 was added to 2500 mL of water. The boiling point of the solution is now 375.25 K. What mass of KMnO4 (in grams) was added to the water? (Kb water = 0.512 \u00c2\u00b0C\/m, MW KMnO4 = 158 g\/mol, density of water = 1.0 g\/mL, boiling point of water = 373.15 K) [&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-16959","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16959","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=16959"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16959\/revisions"}],"predecessor-version":[{"id":16960,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16959\/revisions\/16960"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=16959"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=16959"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=16959"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}