Question

Two elements A and B from compounds having formula `"AB"_(2)` and `"AB"_(4)`. When dissolved in 20g of Benzene `(C_(6)H_(6))`, 1g of `"AB"_(2)` lowers the freezing point by 2.3 K whereas 1.0g of `"AB"_(4)` lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg `"mol"^(-1)`. Calculate atomic masses of A and B.

Answer 1

Calculation of molecular masses of compounds `"AB"_(2)` and `"AB"_(4)`
For `"AB"_(2)` compounds
`m=(K_("f")xx1000xxw)/(DeltaT_("f")xxW)=(5.1xx1000xx1)/(2.3xx20)=110.87" gms"//"mole"`
For `"AB"_(4)` compound
`m=(5.1xx1000xx1)/(1.3xx20)=196.15" gms"//"mole"`
Calculation of the atomic masses of elements
Atomic mass of element A = x
Atomic mass of element B = y
Molecular mass of `"AB"_(2)=x+2y`
Molecular mass of `"AB"_(4)=x+4y`
`a+2b=110.87" "...(1)`
`a+4b=196.15" "...(2)`
Equation (2) - Equation (1)
`x+4y-x-2y`
`196.15-110.87`
`2y=85.28`
`y=42.64`
`x+2y=110.87`
`x+85.28=110.87`
`x=25.59`
Atomic mass of element A = 25.59 u
B = 42.64 u