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Question

# Vapour pressure of solution containing 6g of a non-volatile solute in 180 g of water is 20 torr. If 1 mole water is further added, then vapour pressure increases by 0.02 torr. Which of the following is not correct?

A
The molecular weight of solute is 55 g mol1
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B
The vapour pressure of pure water is 20.22 torr
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C
Addition of more water in the solution will further raise the vapour pressure of the solution
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D
The vapour pressure of pure water is 22.22 torr
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Solution

## The correct option is D The vapour pressure of pure water is 22.22 torrp1= vapour pressure of solution p∘1=vapour pressure of pure solvent According to Raoult's law, p∘1−p1p∘1=w2M1w1M2⇒p∘1−20p∘1=6×18M×180=6×1M×10.......(1) Again by adding one mole of water, we have p∘1−20.02p∘1=6×18M×198=6×1M×11.......(2) Dividing equation (1) by (2), we get p∘1−20p∘1×p∘1p∘1−20.02=6×1M×10×M×116×1⇒p∘1−20p∘1−20.02=1110⇒p∘1=20.22 torr Putting the value of p∘1 in the equation (2), we get M=55.14≈55 On dilution, ΔP decreases. So, vapour pressure increases.

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