Q45 / 97 · NTA Abhyas (JEE Main) 2020 · Chemistry · Solutions

100 g of liquid A (molar mass 140 g  mol −1 ) was dissolved in 1000 g of liquid B (molar mass 180 g  mol −1 ). The vapour pressure of pure liquid B was found to be 500 torr . Calculate vapour pressure of A (in torr) the solution if the total vapour pressure of the solution is 475 torr.

Correct answer: 32.0

Solution

Step I : Calculation of vapour pressure of pure liquid A (p°A)
​Number of moles of liquid A
n A = W A M A = (100 g) 140 g mol - 1 = 0.7143 mol
Number of moles of liquid B
n B = W B M B = (1000 g) 180 g mol - 1 = 5.5556 mol
Mole fraction of A
x A = n A n A + n B
= (0.7143 mol) (0.7143 + 5.5556) mol = 0.7143 6.2699 = 0.1139
Mole fraction of B(x B ) = 1−0.1139=0.8861
Vapour pressure of pure liquid B,  B(p B o ) = 500 torr
Total vapour pressure of solution (p)=475 torr
According to the Raoult's law
p =  p A o   x A  +  p B o   x B
475 torr =  p A o  × (0.1139) + 500 torr × (0.8861)
475 torr =  p A o  × (0.1139) + 443.05 torr
p A o = 475 - 443.05 torr (0.1139) = 31.95 0.1139 torr = 280.5 torr

Step II : Calculation of vapour pressure of A in the solution (pA)
According to Raoult's law,
p A  =  p A o x A  = (280.5 torr) × (0.1139)
p A  = 32.0 torr

Hence, the correct answer is 32.0.

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