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

31.25

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3.125

No worries! We‘ve got your back. Try BYJU‘S free classes today!

312.5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 31.25
$P^{\circ} - P_{S} \propto$ loss in weight of water chamber and $P_{S} \propto$ loss in weight of solution chamber.
$\frac{p^{\circ} - p_{S}}{p_{S}} = \frac{n_{2}}{n_{1}} = \frac{W_{2} \times M w_{1}}{M w_{2} \times W_{1}}$
or $\frac{0.04}{2.50} = \frac{5 \times 18}{M w_{2} \times 180}$
$∴ M w_{2} = 31.25$

Suggest Corrections

Similar questions

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:

Dry air was passed successively through a solution of 5 gm of a solute in 80 gm of water and then through pure water. The loss in weight of solution was 2.50 gm and that of pure solvent 0.04 gm . What is the molecular weight of the solute?

Q. Dry air was passed through a beaker solution containing

15 g

of solute and

120 g

of water. Then, it passed through a beaker containing pure water solvent. Loss in weight of the solution beaker was

2 g

and loss in weight of solvent beaker was

0.05 g

. What is the molecular weight of the solute ?

Q. Dry air is passed through a solution containing 10 g of a solvent in 90 g of water and then through pure water. The loss in weight of the solution is 2.5 g and that of the pure solvent is 0.05 g. Calculate the molecular weight of the solvent.

Q. A current of dry air is passed through a solution of

2.64 g

of non-volatile solute dissolved in

30.0 g

of ether and then through the pure ether. The loss in weight of solution was

0.645 g

and that of the ether was

0.0345 g

. The molecular weight of the solid is :

Join BYJU'S Learning Program

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

The correct option is A 31.25P∘−PS∝ loss in weight of water chamber and PS∝ loss in weight of solution chamber. p∘−pSpS=n2n1=W2×Mw1Mw2×W1 or 0.042.50=5×18Mw2×180 ∴Mw2=31.25