A set of solution is prepared using 180 g of water as a solvent and 10 g of different non - volatile solutes A, B and C . The relative lonwering of vapour pressure in the presence of these solutes are in the order . [ Given , molar mass of `A = 100 g mol^(-1)` `B = 200 g mol^(-1) , C = 10 ,000 g mol^(-1)`]

Options

A. `B gt C gt A`
B. `C gt B gt A`
C. `A gt B gt C`
D. `A gt C gt B`

Answer

C

Explanation

Relative lowering of vapour pressure is equal to mole fraction of solute.
Number of moles of `A=(10)/(100) rArr 10^(-1)` mol , Number of moles of `B=(10)/(200) rArr 5xx10^(-2)` mol
Number of moles of `C=(10)/(10000) rArr 10^(-3)` mol
In same amount of solvent, mole fraction of A is highest and mole fraction of B is higher than C.
So, relative lowering vapour pressure is in order `A gt B gt C`.