1.

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 :

Answer»

`60,90`
`40,40`
`40,10`
`10,40`

SOLUTION :`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`


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