Question

The glucose solution (C6H12O6) in water freezes at a temperature of - 0.18 ° C. If the Kf of water is 1.86 ° C / molal.

Calculate how many grams of glucose dissolved in per liter of water… .. (Ar C = 12 g / mol, Ar H = 1 g / mol, Ar O = 16 g / mol).

Answer 1

Solution.

The depression in the freezing point ΔT_f​=K_f​m

ΔT_f​ = - 0.18 ° C, K_f of water is 1.86 °C/molal.

m = .096 molal

moles of glucose dissolved in per liter of water = .096

molecular weight of glucose = 180 gm

.096 moles of glucose = (180 x 0.096) gm = 17.28 gm (Ans.)