Question

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 :

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

(1) 160 mm of Hg

(2) 168 mm of Hg

(3) 336 mm of Hg

(4) 350 mm of Hg

Answer 1

Correct option is (3) 336 mm of Hg

Given vapour pressure of benzene, P⁰_A = 280 mm of Hg

Vapour pressure of octane, P⁰_B = 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, n_b = 3

n_o = 2

n_total = n_b + n_o = 3 + 2 = 5

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Raoult's law states that at equilibrium,

P_A = P⁰_A x_A

Where P_A is the partial pressure of A

P⁰_A is vapour pressure of pure A at that temperature.

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

Similarly, P_B = P⁰_B x_B

x_A = n_b / n_total = 3/5

x_B = n_o / n_total = 2/5

By Dalton's law, P_S = P⁰_A x_A + P⁰_B x_B

P_S = 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.

Answer 2

Correct answer is (3).