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# 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.

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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.

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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

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