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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 - CBSE (Science) Class 12 - Chemistry

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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?

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Solution 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 Concept: Ideal and Non-ideal Solutions.
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