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A 1.0 g of substance of molecular formul...

A 1.0 g of substance of molecular formula `AB_(2)`when dissoved in 25 g of benzene reduced the freezing point by `1.25^(@)C`, Determine the atomic mass of A and B (`K_(f)` for benzene= 5.1 kg `mol^(-1)`)

Text Solution

Verified by Experts

The correct Answer is:
Atomic mass of A=56.36=Atomic mass of B=35.7
`M_((AB_(2)))=(K_(f)xxW)/(DeltaT_(f)xxW)=((5.1"kg mol"^(-1))xx(1g))/((1.6K)xx(0.025kg))=127.5" g mol"^(-1)`
`M_((AB_(3)))=(K_(f)xxW)/(DeltaT_(f)xxW)=((5.1"kg mol"^(-1))xx(1g))/((1.25K)xx(0.025kg))=163.2" g mol"^(-1)`
Let the atomic mass of element A = a
Let the atimic mass of element B = b
`therefore" "a+2b=127.5`
a+3b=163.2
On subracting eqn. (i) form eqn. (ii)
`3b-2b=163.2-127.5=35.7 or b = 35.7`
`a+bxx(35.7)=127.5 or a=127.5-71.14=56.36`
`DeltaT_(f)=0.201^(@)C=0.201 K, K_(f)=1.86" K kg mol"^(-1), m=0.10"mol"^(-1)kg`
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