1.613 g sample of an unknown nonionizing molecular compound is added to 250.0 g of benzene (C6H6, MW = 78.11 g/mol). The normal freezing point of the solution is 0.17 C lower than the normal freezing point of pure benzene. Based on this information, find the molecular weight of the unknown compound. benzene = Kf = 5.12 kg x Degrees C/mol.
1.613 g sample of an unknown nonionizing molecular compound is added to 250.0 g of benzene (C6H6, MW = 78.11 g/mol). The normal freezing point of the solution is 0.17 C lower than the normal freezing point of pure benzene. Based on this information, find the molecular weight of the unknown compound. benzene = Kf = 5.12 kg x Degrees C/mol.
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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1.613 g sample of an unknown nonionizing molecular compound is added to 250.0 g of benzene
(C6H6, MW = 78.11 g/mol). The normal freezing point of the solution is 0.17 C lower than the normal
freezing point of pure benzene. Based on this information, find the molecular weight of the unknown
compound. benzene = Kf = 5.12 kg x Degrees C/mol.
(C6H6, MW = 78.11 g/mol). The normal freezing point of the solution is 0.17 C lower than the normal
freezing point of pure benzene. Based on this information, find the molecular weight of the unknown
compound. benzene = Kf = 5.12 kg x Degrees C/mol.
Expert Solution
Step 1: Introduction to the given data
The depression in freezing point of Benzene, format('truetype')%3Bfont-weight%3Anormal%3Bfont-style%3Anormal%3B%7D%3C%2Fstyle%3E%3C%2Fdefs%3E%3Ctext%20font-family%3D%22math1ed582716bfb4738ccd92405301%22%20font-size%3D%2216%22%20text-anchor%3D%22middle%22%20x%3D%227.5%22%20y%3D%2216%22%3E%26%23x2206%3B%3C%2Ftext%3E%3C%2Fsvg%3E)
Tf =T0-Tf = 0.17 oC
For benzene, Kf = 1.86 oC.m-1
Mass of the unknown non-ionizing solute = 1.613 g
Mass of the solvent (Benzene) = 250.0 g
Moar mass of the unknown non-ionizing solute =? g
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