Freezing point lauric acid = 44.5 degrees C\u00c3\u201a\u00c2<\/p>\n\n\n\n
Freezing point solution = 40.7 degrees C<\/p>\n\n\n\n
Mass of unknown solid = 0.4156g<\/p>\n\n\n\n
Mass of lauric acid = 3.936g\u00c3\u201a\u00c2<\/p>\n\n\n\n
The freezing point depression constant of lauric acid is 4.40C\/molal<\/p>\n\n\n\n
Calculate\u00c3\u201a\u00c2<\/p>\n\n\n\n
a) Freezing point depression<\/p>\n\n\n\n
b) Molality of unknown (m)<\/p>\n\n\n\n
c) Moles of unknown (mol)<\/p>\n\n\n\n
d) Molar Mass of unknown\u00c3\u201a\u00c2<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n Let’s solve step-by-step.<\/p>\n\n\n\n [ The relationship between freezing point depression and molality is: The molality formula is: The molar mass is given by: a) (\\Delta T_f = 3.8^\\circ \\mathrm{C}) The process involves determining the properties of the solution using the freezing point depression principle. First, the freezing point depression ((\\Delta T_f)) is calculated by subtracting the solution’s freezing point from that of the pure solvent. This gives (\\Delta T_f = 3.8^\\circ \\mathrm{C}).<\/p>\n\n\n\n Using the given freezing point depression constant ((K_f = 4.40 \\, \\mathrm{^\\circ C\/molal})), the molality ((m)) of the solution is calculated. Molality is directly proportional to (\\Delta T_f), so dividing (\\Delta T_f) by (K_f) yields (m = 0.8636 \\, \\mathrm{mol\/kg}).<\/p>\n\n\n\n To find the number of moles of solute ((n_{\\text{solute}})), the definition of molality is used: (m = n_{\\text{solute}} \/ \\text{mass of solvent (kg)}). Rearranging and substituting the solvent’s mass in kilograms gives (n_{\\text{solute}} = 0.003398 \\, \\mathrm{mol}).<\/p>\n\n\n\n Finally, the molar mass of the solute ((M_{\\text{solute}})) is determined using the formula (M_{\\text{solute}} = m_{\\text{solute}} \/ n_{\\text{solute}}). Substituting the solute’s mass and the number of moles provides (M_{\\text{solute}} = 122.3 \\, \\mathrm{g\/mol}).<\/p>\n\n\n\n This value indicates the unknown compound’s molecular weight, derived using colligative properties, which depend solely on particle concentration, not their identity.<\/p>\n\n\n\n <\/p>\n","protected":false},"excerpt":{"rendered":" Freezing point lauric acid = 44.5 degrees C\u00c3\u201a\u00c2 Freezing point solution = 40.7 degrees C Mass of unknown solid = 0.4156g Mass of lauric acid = 3.936g\u00c3\u201a\u00c2 The freezing point depression constant of lauric acid is 4.40C\/molal Calculate\u00c3\u201a\u00c2 a) Freezing point depression b) Molality of unknown (m) c) Moles of unknown (mol) d) Molar Mass […]<\/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-183767","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183767","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=183767"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183767\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=183767"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=183767"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=183767"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nGiven Data:<\/strong><\/h3>\n\n\n\n
\n
\n\n\n\na) Freezing Point Depression ((\\Delta T_f))<\/strong><\/h3>\n\n\n\n
\\Delta T_f = T_f^0 – T_f
]
[
\\Delta T_f = 44.5^\\circ \\mathrm{C} – 40.7^\\circ \\mathrm{C} = 3.8^\\circ \\mathrm{C}
]<\/p>\n\n\n\n
\n\n\n\nb) Molality of the solution ((m))<\/strong><\/h3>\n\n\n\n
[
\\Delta T_f = K_f \\cdot m
]
Rearranging for (m):
[
m = \\frac{\\Delta T_f}{K_f}
]
[
m = \\frac{3.8 \\, \\mathrm{^\\circ C}}{4.40 \\, \\mathrm{^\\circ C\/molal}} = 0.8636 \\, \\mathrm{mol\/kg}
]<\/p>\n\n\n\n
\n\n\n\nc) Moles of solute ((n_{\\text{solute}}))<\/strong><\/h3>\n\n\n\n
[
m = \\frac{n_{\\text{solute}}}{\\text{mass of solvent in kg}}
]
Rearranging for (n_{\\text{solute}}):
[
n_{\\text{solute}} = m \\cdot \\text{mass of solvent in kg}
]
[
\\text{Mass of solvent in kg} = \\frac{3.936}{1000} = 0.003936 \\, \\mathrm{kg}
]
[
n_{\\text{solute}} = 0.8636 \\, \\mathrm{mol\/kg} \\cdot 0.003936 \\, \\mathrm{kg} = 0.003398 \\, \\mathrm{mol}
]<\/p>\n\n\n\n
\n\n\n\nd) Molar Mass of Unknown ((M_{\\text{solute}}))<\/strong><\/h3>\n\n\n\n
[
M_{\\text{solute}} = \\frac{m_{\\text{solute}}}{n_{\\text{solute}}}
]
[
M_{\\text{solute}} = \\frac{0.4156 \\, \\mathrm{g}}{0.003398 \\, \\mathrm{mol}} = 122.3 \\, \\mathrm{g\/mol}
]<\/p>\n\n\n\n
\n\n\n\nSummary of Results:<\/strong><\/h3>\n\n\n\n
b) (m = 0.8636 \\, \\mathrm{mol\/kg})
c) (n_{\\text{solute}} = 0.003398 \\, \\mathrm{mol})
d) (M_{\\text{solute}} = 122.3 \\, \\mathrm{g\/mol})<\/p>\n\n\n\n
\n\n\n\nExplanation (300 words)<\/strong><\/h3>\n\n\n\n