Question

Pinene (C10H16) and terpineol (C10H18O) are two of the many compounds used in perfumes and cosmetics to provide a fresh pine scent. At 69.0°C the pure substances have vapor pressures of 100.3 torr and 9.8 torr, respectively. Assuming 50.0 g for each substance, what is the composition of the vapor in terms of mole fractions above a solution of pinene and terpineol at 69.0°C? Show your work.

Answer 1

Number of moles of Pinene(M. Wt of C10H16 = 136) = weight/molecular weight = 50 g/136 g. mol =0.3676 mol

Number of moles of terpineol(M. Wt of C10H18O = 154) = weight/molecular weight = 50 g/154 g.mol=0.3247 mol

Mole fraction of pinene = 0.3676/(0.3676 +0.3247) = 0.3676/0.6923 =0.5310

Mole fraction of terpineol = 0.3247/(0.3676 +0.3247) =0.4690

Now, lets apply Roult's law

Vapor pressure of solution P = P_pinenepinene + P_terpineol​_terpineol = (100.3 * 0.5310) + (9.8 * 0.4690) = 53.2593 + 4.5962 =57.8555 torr

Now, Lets calculate the mole fractions in vapor

Mole fraction of pinene = (P_pinenepinene)/ P = 53.2593 / 57.8555 =0.9206

Mole fraction of terpineol = (P_terpineol​_terpineol ) /P = 4.5962/ 57.8555 =0.0794