— Analytical Chemistry Chemistry Question
Camphor in Benzene
The vapor pressure of pure benzene (C6H6) is 100 torr at 26.1 o C. Calculate the vapor pressure and the freezing point of a solution containing 24.6 g of camphor (C10H16O) dissolved in 100 cm 3 of benzene. The density of benzene is 0.877 g cm -3. The freezing point and the cryoscopic constant (Kf) of pure benzene are 5.50 o C and 5.12 o C kg mol -1, respectively.
Model Answer
xben = nben / (nben + ncam)
nben = 100 cm 3 × 0.877 g cm -3 × (1 mol / 78.1 g) = 1.12 mol
ncam = 24.6 g × (1 mol / 152.2 g) = 0.162 mol
xben = 1.12 mol / (1.12 mol + 0.162 mol) = 0.874
pben = xben p o ben = 0.874 ×100 torr = 87.4 torr
mben = 100 cm 3 × 0.877 g cm -3 × (1 kg/1000 g) = 0.0877 kg
molality of camphor in solution = 0.162 mol camphor / 0.0877 kg benzene = 1.85 mol kg -1
ΔT = Kf m = 5.12 o C kg mol -1 × 1.85 mol kg -1 = 9.46 o C
Since pure benzene freezes at 5.50 o C, the solution will freeze at –3.96 o C.