📖 generic · CBSE Class 12th English Medium · CHEMISTRY · Page 18question

Example 1.8 · Part 2

Chapter 1: Solutions · CHEMISTRY

, which depend upon the nature of the solvent, can be ascertained from the following relations. K f f fus ×  R M T H ( . ) K b b vap ×  R M T H ( . ) Here the symbols R and M stand for the gas constant and molar mass of the solvent, respectively and T f and T b denote the freezing point and the boiling point of the pure solvent respectively in kelvin.

Further, D fus H and D vap H represent the enthalpies for the fusion and vapourisation of the solvent, respectively. Solvent b. p./K K b /K kg mol - f. p./K K f /K kg mol - Water .

. . . Carbon tetrachloride .

. . . Carbon disulphide .

. . . Diethyl ether .

: Molal Boiling Point Elevation and Freezing Point Depression Constants for Some Solvents Fig. . Level of solution rises in the thistle funnel due to osmosis of solvent. g of ethylene glycol (C H O ) is mixed with g of water.

Calculate (a) the freezing point depression and (b) the freezing point of the solution. Depression in freezing point is related to the molality, therefore, the molality of the solution with respect to ethylene glycol = moles of ethylene glycol mass of water in kilogram Moles of ethylene glycol = g g mol  = . mol Mass of water in kg = 600g 1000g kg  = . kg Hence molality of ethylene glycol = .

mol . kg = . mol kg – Therefore freezing poi nt depression, Ä T f = . K kg mol – × .

mol kg – = . K Freezing point of the aqueous solution = . K – . K = .

K . g of a non-electrolyte solute dissolved in g of benzene

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