Two 5 molal solutions are prepared by dissolving a non-electroylete, non -volatile solute separately in the solvents X and Y. The molecular weights of the solvents are M_(x) and M_(Y), respectively where M_(X)=4/3M_(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 comparision to that of solvent, the value of "m" is :

Two 5 molal solutions are prepared by dissolving a non-electroylete, n

1.	Two 5 molal solutions are prepared by dissolving a non-electroylete, non -volatile solute separately in the solvents X and Y. The molecular weights of the solvents are M_(x) and M_(Y), respectively where M_(X)=4/3M_(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 comparision to that of solvent, the value of "m" is :
Answer» <P>`3/4` `4/3` `1/2` `1/4` Solution :Molality=`("Number of MOLES of SOLUTE")/("Mass of solvent (in kg)")` Number of moles of `=5xx1=5` (in both in solvents) Number of moles of solvent `x=1000/(M_(x))` Number of moles of solvent `Y=1000/(M_(Y))` Relative lowering in vapour pressure is GIVEN as `(p^(@)-p_(s))/(p_(s))=(n_(2))/(n_(1))` `((p^(@)-p_(s))/(p_(s)))_("solution in X")=5/(1000/M)=(5M_(x))/1000` `((p^(@)-p_(s))/(p_(s)))_("solution in Y")=5/(1000/(M_(Y)))=(5M_(x))/1000` According to question `1/mxx(5M)/1000=(5M_(Y))/1000` `1/mxx5xx3/4M_(Y)=M_(Y)` (Given `M_(x)=3/4M_(Y)"":.m=3/4`