# Example-1

Example-1

Two liquids

*A*and*B*form ideal solutions. At 300 K, the vapor pressure of solution containing 1 mol of*A*and 3 mol of*B*is 550 mm Hg. At the same temperature, if one more mole of*B*is added to this solution, the vapor pressure of the solution increases by 10 mm Hg. Determine the vapor pressures of*A*and*B*in their pure states (in mm Hg).- 400, 600
- 500, 500
- 600, 400
- None of these

Solution (A)

Since,

*P*= , we have = 550 mm Hg

= 560 mm Hg

That is, 0.25 + 0.75 = 550 mm Hg

= 560 mm Hg

Solving for we get

= 400 mm Hg and = 600 mm Hg

# Example-2

Example-2

Two beakers

*A*and*B*present in a closed vessel. Beaker*A*contains 15.4 g aqueous solution of urea, containing 18 g of glucose. Both solutions are allowed to attain the equilibrium. Determine weight % of glucose in its solution at equilibrium:- 6.72
- 14.49
- 16.94
- 20

Solution (B)

Mole fraction of urea in its solution =

Mole fraction of glucose =

Since the mole fraction of glucose is less, so vapor pressure above the glucose solution will be higher than the pressure above urea solution. Hence some H

_{2}O molecules will transfer from glucose to urea side in order to make the solutions of equal mole fraction to attain equilibrium. Let*x*mol of H_{2}O transferred. Therefore,Mass of glucose solution = 196.2 â€“ 4 Ã— 18 = 124.2

Weight % of glucose = = 14.49

# Example-3

Example-3

The vapor pressures of two pure liquids

*A*and*B*that form an ideal solution are 100 and 900 torr, respectively, at temperature*T*. This liquid solution of*A*and*B*is composed of 1 mol of*A*and 1 mol of*B*. What will be the pressure, when 1 mol of mixture has been vaporized?- 800 torr
- 500 torr
- 300 torr
- None of these

Solution (C)

Let

*n*mole of_{B}*B*present in 1 mol of mixture that has been vaporized. Thus,*y*=_{B}*n*/1 ._{B}Mole fraction of

*B*in the remaining liquid phase is given byAfter substitution of values of

*x*and_{B}*y*in Eqs. (i) and (ii), we get_{B}or

so

â‡’

â‡’ 300 torr

# Example-4

Example-4

A solution containing 30 g of non-volatile solute in exactly 90 g water has a vapor pressure of 21.85 mm Hg at 25Â°C. Further 18 g of water is then added to the solution. The resulting solution has a vapor pressure of 22.15 mm Hg at 25Â°C. Calculate the molecular weight of the solute.

- 74.2
- 75.6
- 67.83
- 78.7

Solution (C)

We have,

By Eq. (i), we get

By Eq. (ii), we get

0.30

*m*= 20.35# Example-5

Example-5

The boiling point of water (100Â°C) becomes 100.52Â°C if 3 g of a non-volatile solute is dissolved in 200 mL of water. The molecular weight of solute is: (K

_{b}for water is 0.6 K-m)- 12.1 g/mol
- 15.4 g/mol
- 17.3 g/mol
- 20.4 g/mol

Solution (C)

First boiling point of water = 100Â°C

Final boiling point of water = 100.52Â°C