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 :
By BYJU'S Exam Prep
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
Table of content
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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