Question

The vapour pressure of pure liquid (molecular
weight = 50) at 25°C is 640 mm of Hg and the
vapour pressure of a solution of a solute in the
liquid at the same temperature is 600 mm of Hg.
Molality of solution is
(A) 3/4
(B) 3/8
(C)4/3
(D) 4/6​

Answer 1

Answer:

c) 4/3

Explanation:

Kindly check my attached image for clear explanation, my friend.

Answer 2

Given: Vapour pressure of pure liquid - P° = 640mm of Hg

vapor pressure of a solution of the solute in liquid - P = 600mm of Hg

To Find: Molality of the solution - M =?

Solution:

• Molality = moles of solute (n)/ mass of solvent in kg(w₂)

• n = given mass(m')/molecular mass (M')
• According to Raoult's law:-
• P° - P/P° = moles of solute/moles of solvent
• P° - P/P° = $\frac{mass of solute(m_{1} ) * molecular mass of solvent (M'_{2} )}{mass of solvent(m_{2}) * molecular mass of solute(M'_{1}) }$

• P° - P/P° = molality of solution (M) × M'₂ /1000 (to convert to kg)

Applying the above Formula to calculate the molality of solution:-

P° - P/P° = M × M'₂/ 1000

640 - 600/640 = M × 50/1000

40/640 = M × 1/20

1/16 = M × 1/20

M = 20/16

M = 5/4