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Q.

Two 5 molal solutions are prepared by dissolving nonelectrolyte, non-volatile solute seperately in the solvents X and Y. The molecular weights of solvents are M_x and M_y respectively where M_x= $\frac{4}{3} M_{y}$ .The relative lowering of vapour pressure of the solution in "X" is "m" times that of the solution in"Y". Given that the number of moles of solute is very small in comparison to that of solvent, the value of “m” is:

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a

1/4

b

3/4

c

1/2

d

4/3

answer is A.

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Detailed Solution

$M_{x} = \frac{4}{3} M_{y}$

RLVP of ${(\frac{∆ p}{p})}_{x} = m {(\frac{∆ p}{p})}_{y}$

$\frac{∆ p}{p} \infty χ_{s o l u t e}$

5 molal solution means 5 moles of solute is present in 1000 gm of solvent.

moles of solvent = $\frac{1000}{M_{solvent}}$

$χ_{solute} = \frac{5}{5 + \frac{1000}{M}}$

$\frac{1000}{M} > > > 5 ∴ 5 + \frac{1000}{M} \approx \frac{1000}{M}$

$\frac{5}{1000} M_{x} = m \times \frac{5}{1000} M_{y}$

M_x = mM_y

$m = \frac{4}{3}$

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