# The correct option for the value of vapour pressure of a solution at 45°C with benzene to octane in molar ratio 3:2 is :

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Updated on: September 13th, 2023

[At 45°C vapour pressure of benzene is 280 mm of Hg, and that of octane is 420 mm of Hg. Assume Ideal gas]

(a) 350 mm of Hg

(b) 160 mm of Hg

(c) 168 mm of Hg

(d) 336 mm of Hg

Given vapour pressure of benzene, P⁰A = 280 mm of Hg

Vapour pressure of octane, P⁰B = 420 mm of Hg

We have to find the value of vapour pressure of a solution at 45°C with benzene to octane in the given molar ratio.

Also, nb = 3

no = 2

ntotal = nb+no = 3 + 2 = 5

Raoult’s law states that at equilibrium,

PA = P⁰A xA

Where PA is the partial pressure of A

P⁰A is vapour pressure of pure A at that temperature.

xA is the mole fraction of A in the liquid phase.

Similarly, PB = P⁰B XB

xA = nb / ntotal = 3/5

xB = no / ntotal = 2/5

By Dalton’s law, PS = P⁰A xA + P⁰B XB

PS = 280(3/5) + 420(2/5)

= 168 + 168

= 336 mm of Hg

Therefore, the value of vapour pressure of the solution is 336 mm of Hg.

Summary:

## The correct option for the value of vapour pressure of a solution at 45°C with benzene to octane in molar ratio 3:2 is :

The correct option for the value of vapour pressure of a solution at 45°C with benzene to octane in molar ratio 3:2 is 336 mm of Hg.

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