Question

# Pinene (C10H16) and terpineol (C10H18O) are two of the many compounds used in perfumes and cosmetics...

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.

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 = Ppinenepinene + Pterpineolterpineol = (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 = (Ppinenepinene)/ P = 53.2593 / 57.8555 =0.9206

Mole fraction of terpineol = (Pterpineolterpineol ) /P = 4.5962/ 57.8555 =0.0794