SOLUTIONS-2

JEE Advanced 2015 Forum › Physical Chemistry questions › SOLUTIONS-2

(1) 1.5g of a monobasic acid when dissolved in 150g of water lowers the freezing point by 0.165^Â°C. 0.5g of the same acid when titrated,after dissolution in water,requires 37.5ml of N/10 alkali. Calculate the degree of dissociation of the acid (K_f for water = 1.86^Â°C mol^-1).

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2 Answers

govind ·2010-01-19 00:05:52

in this question calculate the number of molecular mass of the acid from the titration part 0.5/M = 37.5/1000 * 1/10
from this u will get M=133g
now calculate the molality of 1.5g acid solution in 150g water
m = 1.5/133 * 1000/150
m = 0.075
now put âˆ‚T = i K_f m
0.165 = i * 1.86 * 0.075
i = 1.18
let the degree of ionisation = x
since it's a mono basic acid so 1+x = i
x = 1.8