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02/10/2020 11:21 am
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The vapour pressure of a solvent decreases by 10 mm of Hg when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be 20 mm of Hg?
(a) 0.8
(b) 0.6
(c) 0.4
(d) 0.2
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02/10/2020 3:19 pm
(b) 0.6
Explanation:
According to Raoult's law
Δp/p° = n/n + N (mole fraction of solute)
10/p° = 0.2 ∴ p° = 50 mm of Hg
For other solution of same solvent
20/p° = n/n + N (Mole fraction of solute)
=> 20/50 = Mole fraction of solute
=> Mole fraction of solute = 0.4
As mole fraction of solute + mole fraction of solvent = 1
Hence, mole fraction of solvent = 1 – 0.4 = 0.6