Two 5 molal solution are prepared by dissoliving a non elecrtroyte non volatile solute separately in the solvents X nad Y The molecular weights of the solvents are `M_(x) and M_(y)` respecitvely where
`M_(x)=3/4 M_(y)` The relative lowering of vapour pressure of the solution in 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 :

Options

A. `3/4`
B. `1/2`
C. `1/4`
D. `4/3`

Answer

A

Explanation

The relationship between molar masses of the two solvents is
`M_X = 3/4 M_Y …(i)`
The relative lowering of vapour pressure of the two solution is
`((DeltaP)/P)_X = m((DeltaP)/P)_Y`...(i)
But, the relative lowering of vapour pressure of solutions is directly proportional to the mole fraction of solute.
Given 5 molal solution, means 5 moles of solute are dissolved in1 kg (or 1000g) of solvent.
The number of moles of solvent `=(1000g)/M`
The mole fractionof solute `= 5/(1000//M)`
`=M xx 5/1000`
hence `M_X xx5/1000 = m xx M_Y xx 5/1000`...(ii)
Substitute equation (i) in equation (ii)
`3/4xx M_Y xx 5/1000= m xx M_Y xx 5/1000`
`m = 3/4`