Two elements `A` and `B` form compounds having molecular formula `AB_(2)` and `AB_(4)`. When dissolved in `20 g` of benzene, `1 g`of `AB_(2)` lowers the freezing point by `2.3 K`, whereas `1.0 g` of `AB_(4)` lowers it by `1.3 K`. The molar depression constant for benzene is `5.1 K kg mol^(-1)`. Calculate the atomic mass of `A` and `B`.

Explanation

`M_(B)=(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))`
Calculation of the molecular masses of sdthe two compounds,
`"For the coimpound"AB_(2)`,
`W_(B)=1g,W_(A)=20g=0.02 kg, DeltaT_(f)=2.3 K,`
`K_(f)=5.1 K kg mol^(-1)`
`M_(B)=((5.1 K kg mol^(-1))xx(1g))/((2.3K)xx(0.02 kg))=11087 g mol^(-1)`
`"For the compound"AB_(4)`
`W_(B)=1g,W_(A)=20g0.02 kg, DeltaT_(f)=1.3 K,`
`K_(f)=5.1 K kg mol^(-1)`
`M_(B)=((5.1K kg mol^(-1)))/((1.3K)xx(0.02 kg))=11087 g mol^(-1)`
`"Calculation of the atimic masses os elements A and B."`
`" Let the atimic mass of element A=a "`
`"Let the atimic mass of element B = b "`
`"Molecular mass of "AB_(2)=a+2b`
`"Molecular mass of "AB_(4)=a+4b`
`"According to available information. "`
a+2b=110.87
a+4b=196.15
`"Substiting the value of b in eqn. (i)"`
2b=(196.15-110.87)=85.28 or b=85.28/2 = 42.64
`" Substituting the value of b in eqn. (i)"`
a+2(42.64)=110.87 or a =11087 - 85.28=25.59
`" Thus, Atomic mass of element "A= 25.59 or 25.59 u `
`"Atomic mass of element "A=342.64 or 42.64 u`