Home
Class 12
CHEMISTRY
Dry air is passed through a solution co...

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

A

50

B

180

C

100

D

25

Text Solution

Verified by Experts

The correct Answer is:
C
According to Raoult.s law
` (p-ps)/(p)= (n)/(n+N) = (0.05)/(2.5+ 0.05)`
`= (0.05)/(2.55) = (1)/(51)`
Weight of solute = ` (w)/(W) xx M xx (p)/(p-ps)`
` " " = (10 xx 18)/(90 ) xx 51`
` " " =102 g`
Promotional Banner

Similar Questions

Explore conceptually related problems

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 the weight of solution was 2.50 g and that of pure solvent 0.04 g. The molecular weight of the solute is.

Calculate the mass percent of a solution containing 2.5g of KCl in 50 g of water.

A given solution of NaOH contains 2.00 g of NaOH per litre of solution. Calculate the molarity of this solution?

A solution is prepared by adding 2 g of a substance A to 18 g of water. Calculate the massper cent of the solute

Calculate the molality of a solution of NaOH containing 20g of NaOH in 400 g solvent.

The density of 2.5 M acetic acid in water is 1.02g ml^(-1) . Calculate the molality of the solution.

Calculate the mole fraction of NaOH in a solution containing 20 g of NaOH per 360 g of water.

A 6.9 M solution of KOH in water contains '30 %' by weight of 'KOH'. Calculate the density of the solution?