Ratio at `DeltaT_(b)//K_(b)` of `6% AB_(2)` and `9%A_(2)B(AB_(2)` and A_(2)B` both are non-electolytes) is `1 "mol"//kg` in both cases. Hence, atomic masses of `A` and `B` are respectivly :

Options

A. `60,90`
B. `40,40`
C. `40,10`
D. `10,40`

Answer

C

Explanation

`DeltaT=(1000K_(b)w_(1))/(m_(1)w_(2))`
`(DeltaT)/(K_(b))(AB_(2)=(1000xx6)/(m_(1)xx100)=1`
`therefore m_(1)(AB_(2))=60=A+2B`
`(DeltaT)/(K_(b)_(A_(2)B)=(1000xx9)/(m_(1)xx100)=1`
`therefore m_(1)(A_(2)B)=90=2A+B`
`therefore A=40,B=10`