Question

Calculate the vapor pressure in torr of a solution that is made by dissolving 59.51 grams of glucose (C6H12O6) in 112.2 mL of water at 26.7 oC. The vapor pressure of pure water at 26.7 oC is 26.271 torr and the density of water at that temperature is 0.99669 g/mL.

Answer 1

Mass of water = density × volume = 0.99669 × 112.2 = 111.83g

Number of moles of glucose = 59.51/180 = 0.3306

Number of moles of water = 111.83/ 18 = 6.213

Mole fraction of the solution = 0.3306/(0.3306+6.212) = 0.05

P^o- P/(P^o) = X

P= 26.271 - (26.271×0.05)

P = 24.96 torr