Benzene and toluene form ideal solutions over the entire range of composition. The vapour pressure of pure benzene and toluene 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 toluene.
Molar mass of benzene (C6H6) = 6 x 12 + 6 x1
= 78 g mol-1
Molar mass of toluene (C6H5CH3) = 7 x 12 + 8 x 1
= 92 g mol-1
Now, no. of moles present in 80 g of benzene = \(\frac{80}{78}\)mol
= 1.026 mol
And, no. of moles present in 100 g of toluene = \(\frac{100}{92}\) mol
= 1.087 mol
∴ Mole fraction of benzene, xb = \(\frac{1.026}{1.026 + 1.087}\) = 0.486
And, mole fraction of toluene, xt = 1 - 0.486 = 0.514
It is given that vapour pressure of pure benzene, pb0 = 50.71 mm Hg
And, vapour pressure of pure toluene, pt0 = 32.06 mm Hg
Therefore, partial vapour pressure of benzene, pb = xb x pb
= 0.486 x 50.71
= 24.645 mm Hg
And, partial vapour pressure of toluene, pt = xt x pt
= 0.514 x 32.06
= 16.479 mm Hg
Hence, mole fraction of benzene in vapour phase is given by:
= \(\frac{p_b}{p_b + p_t}\)
= \(\frac{24.645}{24.645 + 16.479}\)
= \(\frac{24.645}{41.124}\)
= 0.599
= 0.6