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A solution contains 6 g of a non-volatile solute dissolved in 180 g of water, and has a vapour pressure of 20 torr. When 1 mole of water is added to this solution the vapour pressure increases by 0.02 torr. Find the vapour pressure of pure water.

1. 20.22 torr
2. 20.02 torr
3. 19.78 torr
4. 19.88 torr

Correct answer: 20.22 torr

Solution

Let moles of solute = n, P0 = vapour pressure of pure water. Raoult: P = P0 * (moles water)/(moles water + n). Equation 1: 20 = P0*10/(10+n). Equation 2: 20.02 = P0*11/(11+n). Dividing and solving gives n = 1/9 mol. Substituting back: P0 = 20*(10+1/9)/10 = 20*91/90 = 1820/90 approx 20.22 torr.

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