vapour pressure of a substance above a solution is proportional to the partial fraction (of total moles) of the substance in the solution. 180 gm = 10 moles of water. Let M be molar mass of solute. Let k be proportionality constant in Torr.
10/[10+6/M] = k 20 => 10 = k 20 [10+6/M]
11/[11+6/M] = k 20.02 => 11 = k 20.02 [11+6/M]
200.2*11 + 200.2*6/M = 2200 + 220*6/M
2.2 = 19.8*6/M => M = 54
4.5 / 91 = k => 1/k = 910/45 = 20.2222
for pure water the vapoour pressure will be 20.22 torr...
Another way, when mole fraction of water is increased by 0.001 from 0.989 to 0.990 above, the vapor pressure increased by 0.02 torr. If mole fraction is increased by 0.01 to make that pure water, the pressure increases by 0.02*10 = 0.2 torr. So pressure will be 20.2 torr
