Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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Benzene and toluene form ideal solution 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 - Chemistry

Question

Benzene and toluene form ideal solution 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.

Solution

Molar mass of benzene  (C6H6) = 6 x 12 + 6 x 1

= 78 gmol-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  = 80/78 mol = 1.026 mol

And, no. of moles present in 100 g of toluene = 100/92 mol = 1.087 mol

∴Mole fraction of benzene, x_b = (1.026)/(1.026 + 1.087) = 0.486

And, mole fraction of toluene,  x_1 = 1 - 0.486 = 0.514

It is given that vapour pressure of pure benzene, p_b^0 = 50.71 mm Hg

And, vapour pressure of pure toluene, p_1^0 = 32.06 mm Hg

Therefore, partial vapour pressure of benzene, p_b = x_b xx p_b

=0.486 x 50.71

= 24.645 mm  Hg

And, partial vapour pressure of toluene, p_t = x_t xx p_t

= 0.514 x 32.06

= 16.479 mm Hg

Hence, mole fraction of benzene in vapour phase is given by:

p_b/(p_b + p_t)

= 24.645/(24.645+ 16.479)

= 24.645/41.124

= 0.599

= 0.6

Is there an error in this question or solution?