The mole fraction of a solute in a solutions is `0.1`. At `298K` molarity of this solution is the same as its molality. Density of this solution at 298 K is `2.0 g cm^(-3)`. The ratio of the molecular weights of the solute and solvent, `(MW_("solute"))/(MW_("solvent"))` is

Answer

9

Explanation

`m=(X_(A)xx1000)/(X_(B)xxM_(A))`
`m=(1000)/(9M_(A))`……(i)
`M=(n_(B)xx1000xxd)/(n_(A)xxM_(A)+n_(B)xxM_(B))=(X_(B)xx1000xx d)/(X_(A)xx M_(A)+X_(B)xxM_(B))`
`=(200)/(0.9M_(A)+0.1M_(B))=(2000)/(9M_(A)+M_(B))` ……(ii)
As m = M
`(1000)/(9M_(A))=(2000)/(9M_(A)+M_(B))rArr 9M_(A)+M_(B)=18M_(A)`
`therefore 9M_(A)=M_(B) therefore (M_(B))/(M_(A))=9`