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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).
  1. 400, 600
  2. 500, 500
  3. 600, 400
  4. None of these
Solution (A)
Since, P = Description: 45421.png, we have
 
Description: 45429.png = 550 mm Hg
 
Description: 45437.png = 560 mm Hg
 
That is, 0.25Description: 45445.png + 0.75Description: 45453.png = 550 mm Hg
 
Description: 45465.png = 560 mm Hg
 
Solving for Description: 45473.png we get
 
Description: 45520.png = 400 mm Hg and Description: 45528.png = 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:
  1. 6.72
  2. 14.49
  3. 16.94
  4. 20
Solution (B)
Mole fraction of urea in its solution = Description: 45504.png
Mole fraction of glucose = Description: 45517.png
 
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 H2O 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 H2O transferred. Therefore,
 
Description: 45543.png
 
Mass of glucose solution = 196.2 – 4 × 18 = 124.2
Weight % of glucose = Description: 45551.png = 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?
  1. 800 torr
  2. 500 torr
  3. 300 torr
  4. None of these
Solution (C)
Let nB mole of B present in 1 mol of mixture that has been vaporized. Thus, yB = nB/1 .
Mole fraction of B in the remaining liquid phase is given by
 
 
After substitution of values of xB and yB in Eqs. (i) and (ii), we get
 
or Description: 45622.png
 
so Description: 45630.png
 
⇒ Description: 45641.png
 
⇒ 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 25°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 25°C. Calculate the molecular weight of the solute.
  1. 74.2
  2. 75.6
  3. 67.83
  4. 78.7
Solution (C)
We have,
 
By Eq. (i), we get
 
Description: 45674.png
 
By Eq. (ii), we get
 
Description: 45685.png
 
0.30m = 20.35
 
Description: 45693.png
 

Example-5

Example-5
The boiling point of water (100°C) becomes 100.52°C if 3 g of a non-volatile solute is dissolved in 200 mL of water. The molecular weight of solute is: (Kb for water is 0.6 K-m)
  1. 12.1 g/mol
  2. 15.4 g/mol
  3. 17.3 g/mol
  4. 20.4 g/mol
Solution (C)
First boiling point of water = 100°C
 
Final boiling point of water = 100.52°C
 
Description: 45704.png
 
Description: 45714.png
 
Description: 45723.png Description: 45731.png
 




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