A very small amount of a non-volatile solute (non-associative, non-dissociative) is dissolved in 100 cm 3 of a solvent. At room temperature, vapour pressure of this solution is 98.7 mm of Hg while that of pure solvent is 100 mm of Hg. If the freezing temperature of this solution is 0.72 K lower than that of pure solvent, what is the value of cryoscopic constant of solvent (in K Kg/mol) ? Round off your answer to the nearest whole number. Report your answer as 0 (zero) if you find data insufficient.Given : Molar mass of solvent = 78 g/mol.
Text Solution
Verified by Experts4
(4)
Sol. 
= X solute
∴ 
= X solute = 0.013
Now, Δ T f = K f × m
0.72 = K f × 
∴ K f ≈ 4.2 K Kg/mol
∴ Reported answer = 4
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