Two liquid A and B form ideal solutions. At 300 K,

Question

Two liquid A and B form ideal solutions. At 300 K, the vapour pressure of a solution containing 1 mole of A and 3 moles of B is 550 mm of Hg. At the same temperature, if one more mole of B is added to this solution, the vapour pressure of the solution increases by 10 mm of Hg. Determine the vapour pressure of A and B in their pure states.

Deepak Patra · Accepted Answer

Hello Student,

Please find the answer to your question

P_T = p^o₁ + p^o₂ x₂

At 300 K, the vapour pressure of the solution containing 1 mole of A & 3 moles of B = 550 mm of Hg and vapour pressure of the solution containing 1 mole of A & 4 moles of B at 300 K = 560 mm Hg.

Let the vapour pressure of pure A = p₁^o

And the vapour pressure of pure B = p₂^o

Further, let x₁ and x₂ be the mole fractions of A and B in the solutions. Then the total vapour pressure of solution

P_total = p₁⁰x₁ + p2⁰x₂ …(i)

In solution 1, p_total = 550 mm,

∴ x₁ = 1/1 – 3, x₂ = 3/1 +3 [∵ moles of A = 1 moles of B = 3]

Substituting these values in (i) we get

550 = p₁⁰ * ¼ + p₂⁰ * ¾

Or 550 = p₁⁰/4 + 3p₂⁰/4

2200 = p₁⁰+ 3p₂⁰ …(ii)

In solution 2, p_total = 560 mm,

X₁ 1/1 + 4, x₂ = 4/1 + 4 [∵ moles of A = 1 moles of B = 4]

Substituting the various values, in equ. (i) we get

560 = p₁⁰ * 1/5 + p₂⁰ * 4/5

Or 560 p₁⁰/5 + 4p₂⁰/5

2800 = p₁⁰ + 4p₂⁰ ….(iii)

Solving equation (ii) and (iii), we get

P₂⁰ = 6400 mm of Hg

p₁⁰ = 400 mm of Hg

∴ Vapour pressure of pure A = 400 mm of Hg

And vapour pressure of pure B = 600 mm of Hg

Thanks
Deepak patra
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