QUESTION 2.38

Benzene and naphthalene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and naphthalene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of naphthalene.

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Solution

Molar mass of benzene $(C_{6} H_{6}) = 78 g m o l^{- 1}$
Molar mass of naphthalene $(C_{10} H_{8}) = 128 g m o l^{- 1}$
$n c_{6} H_{6}$ (Number of moles of benzene)
$= \frac{80 g}{(78 g m o l^{- 1})} = 1.026 m o l$
$n_{C_{10} H_{8}}$ (Number of moles of naphthalene)
$= \frac{(100 g)}{(128 g m o l^{- 1})} = 0.781 m o l$
Mole fraction of benzene
$(x_{C_{6} H_{6}}) = \frac{(1.026 m o l)}{(1.026 + 0.781) m o l} = 0.568$
Mole fraction of naphthalene $(x_{C_{10} H_{8}}) = 1 - 0.568 = 0.432$ Applying Raoult's law
Partial vapour pressure of benzene in solution $(p_{C_{6} H_{6}})$
$= p_{C_{6} H_{6}}^{\circ} \times x_{C_{8} H_{6}} = (50.71 m m) \times (0.568) = 28.80 m m$
Partial vapour pressure of naphthalene in solution
$(p_{C_{10} H_{6}}) = p_{C_{10} H_{8}}^{\circ} \times x_{C_{10} H_{8}} = (32.06 m m) \times (0.432) = 13.85 m m$
Total vapour pressure of solution (p) = (28.80 + 13.85) mm
= 42.65 mm
Mole fraction of benzene in vapour phase
$= \frac{x_{C_{6} H_{6}} \times p_{C_{6} H_{6}}^{\circ}}{p_{t o t a l}} = \frac{0.568 \times (50.71 m m)}{(42.65 m m)} = 0.675$
Mole fraction naphthalene in vapour phase
$= \frac{x_{C_{10} H_{8}} \times p_{10} H_{6}^{\circ}}{p_{t o t a l}} = \frac{0.432 \times (32.06 m m)}{(42.65 m m)} = 0.325$

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