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# Engineering

Question

Asked in: JEE Main-2018

Two 5 molal solutions are prepared by dissolving a non-electrolyte non-volatile solute separately in the solvents X and Y. The molecular weights of the solvents are MX and MY , respectively where
M_{X} = \frac{3}{4}M_{Y}. The relative lowering of vapour pressure of the solution in 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 solvent, the value of “m” is :

A.

\frac{4}{}3

B.

\frac{3}4

C.

\frac{1}{}2

D.

\frac{1}{}4

Answers (1)
P Pankaj Sanodiya

 

The relationship between molar masses of two solvents is 

M_X=\frac{3}{4}M_Y...........(1)

 The relative lowering of the vapor pressure of two solutions is But, the relative lowering of the vapor pressure of the solution is directly proportional to the mole fraction of solute. 

M_X\times\frac{5}{1000}=m\times M_Y\times \frac{5}{1000}...........(2)

Substitute equation (1) in equation (2). 

 \frac{3}{4}\times M_Y\times\frac{5}{1000}=m\times M_Y\times \frac{5}{1000}

m=\frac{3}{4}

 

 

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