The vapour pressure of pure water at 26^oC is 25.21 torr. What is the vapour pressure of a solution which contain 20g glucose , C₆H₁₂O₆ , in 70g water?

First calculate the number of moles of water and glucose.

no. of moles of water = n_w = mass of water / molar mass of water = 70 g / 18 g mol^-1 = 3.9 mol

no. of moles of glucose= n_g = mass of glucose / molar mass of glucose = 20 g / 180 g mol^-1 = 0.11 mol

Total number of moles of mixture(n) = 3.9 + 0.11 = 4.0 mol

Now calculate the mole fractions.

mole fraction of water = X_w = n_w / n = 3.9 / 4.0 = 0.97

Now calculate vapor pressures of solution.

p = p_w^oX_w = 25.21 x 0.97 = 24.45 torr