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The vapour pressure of a solvent decreased by 10 mm Hg when a non-volatile solute was added to the solvent. The mole fraction of solute in solution is 0.2, what would be mole fraction of the solvent in decrease in vapour pressure is 20 mm of 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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Explanation:

When a non-volatile solute is added to a solvent, it lowers the vapor pressure of the solvent. This is known as Raoult's law. According to Raoult's law, the vapor pressure of a solvent in a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent.

Let's assume the initial vapor pressure of the pure solvent is P°. When the solute is added, the vapor pressure of the solvent decreases by 10 mm Hg. Therefore, the new vapor pressure of the solvent is P° - 10 mm Hg.

Using Raoult's law:

According to Raoult's law, the vapor pressure of the solvent in the solution can be represented as:

P(solvent) = X(solvent) * P°(solvent)

Where P(solvent) is the vapor pressure of the solvent in the solution, X(solvent) is the mole fraction of the solvent in the solution, and P°(solvent) is the vapor pressure of the pure solvent.

Given that the mole fraction of the solute in the solution is 0.2, we can represent the mole fraction of the solvent as:

X(solvent) = 1 - X(solute)
X(solvent) = 1 - 0.2
X(solvent) = 0.8

Now, we can use the equation P(solvent) = X(solvent) * P°(solvent) to find the vapor pressure of the solvent in the solution:

P(solvent) = 0.8 * P°(solvent)

Since the decrease in vapor pressure is 20 mm Hg, we can write the equation as:

P°(solvent) - 10 = 0.8 * P°(solvent)

Simplifying the equation:

0.2 * P°(solvent) = 10
P°(solvent) = 10 / 0.2
P°(solvent) = 50 mm Hg

Now, we can substitute this value back into the equation P(solvent) = 0.8 * P°(solvent) to find the mole fraction of the solvent:

P(solvent) = 0.8 * 50
P(solvent) = 40 mm Hg

Therefore, the mole fraction of the solvent in the solution when the decrease in vapor pressure is 20 mm Hg is 0.6 (option B).
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The vapour pressure of a solvent decreased by 10 mm Hg when a non-volatile solute was added to the solvent. The mole fraction of solute in solution is 0.2, what would be mole fraction of the solvent in decrease in vapour pressure is 20 mm of Hg:a)0.8b)0.6c)0.4d)0.2Correct answer is option 'B'. Can you explain this answer?
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