Dry air was passed successively through a solution of 5 g of a solute in 180 g of water and then through pure water. The loss in weight of solution was 2.50 g and that of pure solvent 0.04 g. The molecular weight of the solute is:

Question

Dry air was passed successively through a solution of 5 g of a solute in 180 g of water and then through pure water. The loss in weight of solution was 2.50 g and that of pure solvent 0.04 g. The molecular weight of the solute is: (A) 31.25 (B) 3.125 (C) 312.5 (D) None of these

Tardigrade Admin · Accepted Answer

P0-Ps propto loss in weight of water chamber and Ps propto loss in wt. of solution chamber (Po-Ps/Po)=(n/N)=(w× M/m× W) or (0.04/2.50)=(5×18/m×180) ∴ m=31.25

Q. Dry air was passed successively through a solution of 5 g of a solute in 180 g of water and then through pure water. The loss in weight of solution was 2.50 g and that of pure solvent 0.04 g. The molecular weight of the solute is:

$31.25$

$3.125$

$312.5$

None of these

$P_{0} - P_{s} \propto$ loss in weight of water chamber and $P_{s} \propto$ loss in wt. of solution chamber
$\frac{P ^{o} - P _{s}}{P ^{o}} = \frac{n}{N} = \frac{w \times M}{m \times W}$
or $\frac{0.04}{2.50} = \frac{5 \times 18}{m \times 180}$
$∴ m = 31.25$

Q. Dry air was passed successively through a solution of 5 g of a solute in 180 g of water and then through pure water. The loss in weight of solution was 2.50 g and that of pure solvent 0.04 g. The molecular weight of the solute is:

31.25

3.125

312.5

None of these

P0​−Ps​∝ loss in weight of water chamber and Ps​∝ loss in wt. of solution chamber PoPo−Ps​​=Nn​=m×Ww×M​ or 2.500.04​=m×1805×18​ ∴m=31.25

$31.25$

$3.125$

$312.5$

$P_{0} - P_{s} \propto$ loss in weight of water chamber and $P_{s} \propto$ loss in wt. of solution chamber
$\frac{P ^{o} - P _{s}}{P ^{o}} = \frac{n}{N} = \frac{w \times M}{m \times W}$
or $\frac{0.04}{2.50} = \frac{5 \times 18}{m \times 180}$
$∴ m = 31.25$