Question

The vapour pressure of pure benzene at 25∘C is 639.7mm of mercury and the vapour pressure of a solution of a solute in benzene at the same temperature is 631.9mm of mercury.Calculate the molality of the solution.

Answer 1

Here given in the question :-

The pressure of benzene, p = 639.7 and pₓ = 631.9 

Assume that the n amount moles dissolved in 1000 gm of Benzene .
Since we know the molar mass of benzene (C₆H₆) = 78
The number of moles of Solvent.
N = 1000/78 = 12.82

Therefore applying the formula .

$(p - p_{x} ) /p = n/N$
(639.7 -631.9) / 639.7 = n/12.82
7.8/639.7 = n/12.82
n = (7.8 × 12.82) / 639.7
n = 99.99/ 639.7
n = 0.156



Hope it Helps . :-)

Answer 2

p = 639.7 and pₓ = 631.9

Assume that the n amount moles dissolved in 1000 gm of Benzene .

Since we know the molar mass of benzene (C₆H₆) = 78

The number of moles of Solvent.

N = 1000/78 = 12.82

Therefore applying the formula .

(p - p_{x} ) /p = n/N(p−p

x

)/p=n/N

(639.7 -631.9) / 639.7 = n/12.82

7.8/639.7 = n/12.82

n = (7.8 × 12.82) / 639.7

n = 99.99/ 639.7

n = 0.156