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Question: Two 5 molal solutions are prepared by dissolving a non-electrolyte non-volatile solute separately in...

Two 5 molal solutions are prepared by dissolving a non-electrolyte non-volatile solute separately in the solvents X and Y. The molecular weight of the solvents are MXMY{M}_{X}{M}_{Y} respectively where MX=34MY{M}_{X} = \cfrac {3}{4} {M}_{Y}. The relative lowering of the vapour pressure of the solution X is "m" times that of the solution in Y. Given that the number of moles of solute is very small in comparison to that of the solvent, the value of "m" is?
A. 34\cfrac {3}{4}
B. 12\cfrac {1}{2}
C. 14\cfrac {1}{4}
D. 43\cfrac {4}{3}

Explanation

Solution

When a solute is added to the solvent the vapour pressure of the solvent decreases. This is known as the lowering of vapour pressure. The lowering of the vapour pressure depends upon the amount of non-volatile solute is added in the solution.

Complete step by step answer: It is given is the question that two solutions are prepared by dissolving a non-electrolyte non-volatile solute separately in the solvents X and Y.
It is also given that MX=34MY { M }_{ X } = \cfrac { 3 }{ 4 } { M }_{ Y } ------ (1)
We know that the relative lowering of the vapour pressure of two solutions is given as:
(ΔPP)X=m(ΔPP)Y{ (\cfrac { \Delta P }{ P } ) }_{ X } = m{ (\cfrac { \Delta P }{ P } ) }_{ Y }
The above relation can be written because it is given that the relative lowering of the vapour pressure of the solution X in "m" times that of the solution in Y.
Now, we know that the relative lowering of the vapour pressure of a solution is directly proportional to the mole fraction of the solution. Thus, from the vapour pressure relation, we get
Mx×51000=m×MY×51000{ M }_{ x } \times \cfrac { 5 }{ 1000 } = m \times { M }_{ Y } \times \cfrac { 5 }{ 1000 }
Here, 51000\cfrac {5}{1000} gets cancelled from both the sides. And then substituting this equation in equation (1), we get
34×MY=m×MY\cfrac { 3 }{ 4 } \times { M }_{ Y } = m \times { M }_{ Y }
Here, MY{M}_{Y} gets cancelled from both the sides.
m=34m = \cfrac { 3 }{ 4 }

Therefore, the value of m is 34\cfrac {3}{4}. Hence, the correct answer is option (A).

Note: 5 molal solutions of both X and Y are prepared. It means that 5 moles of the solute is dissolved in 1 kg (or 1000g) of the solvent.
Therefore, no. of moles of solvent becomes 1000M\cfrac { 1000 }{ M }
Thus, the mole fraction of the solvent becomes 51000M\cfrac { 5 }{ \cfrac { 1000 }{ M } }