Question

We have taken a saturated solution of $AgBr$. $K _{ sp }$ of $AgBr$ is $12 \times 10^{-14}$. If $10^{-7}$ mole of $AgNO _{3}$ are added to 1 litre of this solution then the conductivity of this solution in terms of $10^{-7} Sm ^{-1}$ units will be

[Given$ \lambda_{\left( Ag ^{+}\right)}^{0}=4 \times 10^{-3} Sm ^{2} mol ^{-1}\lambda_{\left( Br ^{-}\right)}^{0}=6 \times 10^{-3} S m ^{2} mol ^{-1}, \lambda_{\left( NO _{3}^{-}\right)}^{0}=5 \times 10^{-3} Sm ^{2} mol ^{-1}$ ]

(A) 39

(B) 55

(C) 15

(D) 41

Answer 1

The conductivity of this solution in terms of 10^–7Sm^–1 units will be 39. So, The correct option of this question is (A).