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The vapour pressure at a given temperature of an ideal solution containing 0.2 mol of a non-volatile solute and 0.8 mol of solvent is 60 mm of Hg. The vapour pressure of the pure solvent at the same temperature will be
  • a)
    120 mm of Hg
  • b)
    150 mm of Hg
  • c)
    60 mm of Hg
  • d)
    75 mm of Hg
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
The vapour pressure at a given temperature of an ideal solution contai...
Given:

Moles of non-volatile solute (n₁) = 0.2 mol

Moles of solvent (n₂) = 0.8 mol

Vapour pressure of solution (P) = 60 mm of Hg

To find:

Vapour pressure of the pure solvent (P₀)

Explanation:

According to Raoult's law,

P = P₀X₂

where P₀ is the vapour pressure of the pure solvent, X₂ is the mole fraction of the solvent.

The mole fraction of the solvent is given by,

X₂ = n₂/(n₁ + n₂)

Substituting the given values,

X₂ = 0.8/(0.2 + 0.8) = 0.8

Now,

P = P₀X₂

P₀ = P/X₂ = 60/0.8 = 75 mm of Hg

Therefore, the vapour pressure of the pure solvent at the given temperature is 75 mm of Hg.

Hence, the correct option is (d) 75 mm of Hg.
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