Question

Calculate the mole fractions of benzene and naphthalene in the vapour phase when an ideal liquid solution is formed by mixing 128 g of naphthalene with 39g of benzene. It is given that the vapour pressure of pure benzene is 50.71 mm Hg and the vapour pressure of pure naphthalene is 32.06 mm Hg at 300 K.

Answer 1

P⁰_{Pure benzene} = 50.71 mm Hg 

P⁰_nepthalene = 32.06 mm Hg

Number of moles of benzene = \(\frac{39}{78}\) = 0.5 mol

Number of moles of naphthalcne = \(\frac{128}{128}\) = 1 mol 

Mole fraction of benzene = \(\frac{0.5}{1.5}\) = 0.33 

Mole fraction of naphthalene = 1 – 0.33 = 0.67

Partial vapour pressure of benzene =P⁰_benzene x (Mole fraction of benzene)

= 50.71 x 0.33 = 16.73 mm Hg

Partial vapour pressure of naphthalene = 32.06 x 0.67 = 21.48mm Hg

Mole fraction of benzene in vapour phase = \(\frac{16.73}{16.73+21.48}\) = \(\frac{16.73}{38}\) = 0.44

Mole fraction of naphthalene in vapour phase = 1 – 0.44 = 0.56