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The vapour pressure of a solvent decreased by 10mmHg when a nonvolatile solute was added to the solvent. The mole fraction of the solute in the solution
is 0.2. What would be the mole fraction of the solvent if the decrease in the vapour pressure is to be 20 mm Hg
  • a)
    0.8
  • b)
    0.6
  • c)
    0.4
  • d)
    0.2
Correct answer is option 'B'. Can you explain this answer?
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Given information:
- The vapour pressure of a solvent decreased by 10 mmHg when a nonvolatile solute was added to the solvent.
- The mole fraction of the solute in the solution is 0.2.
- We need to find the mole fraction of the solvent if the decrease in the vapour pressure is to be 20 mmHg.

Formula:
Raoult's law states that the vapour pressure of a solution is directly proportional to the mole fraction of the solvent present in the solution.

Calculation:
Let P1 be the original vapour pressure of the solvent and P2 be the vapour pressure of the solution after adding the solute.

According to Raoult's law:
P2 = (1-0.2)P1 = 0.8P1

Given that P1 - P2 = 10 mmHg, we have
P1 - 0.8P1 = 10 mmHg
0.2P1 = 10 mmHg
P1 = 50 mmHg

Now we need to find the mole fraction of the solvent when P1 - P2 = 20 mmHg. Let P3 be the new vapour pressure.

According to Raoult's law:
P3 = (1-x)P1

We know that P1 - P3 = 20 mmHg, so
P3 = P1 - 20 mmHg = 30 mmHg

Substituting the values in the equation, we get:
30 mmHg = (1-x)50 mmHg
1-x = 0.6
x = 0.4

Therefore, the mole fraction of the solvent is 0.4.

Answer: Option B.
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