Question

$1 \times 10^{-5} M \,AgNO _3$ is added to $1\, L$ of saturated solution of $AgBr$ The conductivity of this solution at $298 \,K$ is ___ [Given : $K _{\text {SP }}( AgBr )=49 \times 10^{-13}$ at $298\, K$ $\lambda_{ Ag ^{+}}^0=6 \times 10^{-3} S\, m ^2\, mol ^{-1}$ $\lambda_{ Br ^{-}}^0=8 \times 10^{-3} S\, m ^2\, mol ^{-1}$ $\lambda_{ NO _3^{-}}^0=7 \times 10^{-3} \,S\, m ^2 \,mol ^{-1}$ ]

Answer 1

The correct answer is 14.
[Ag+]=10−5
[NO3−​]=10−5
Λm​=1000×Mk​
For Ag+
6×10−3=1000×10−5KAg+​​
KAg+​=6×10−5
⇒6000×10−8
for Br−
8×10−3=1000×4.9×10−8KBr−​​
KBr−​=39.2×10−8
for NO3−​
7×10−3=1000×10−5KNO3−​​​
KNO3​​=7×10−5
=7000×10−8
Conductivity of solution
⇒(6000+7000+39.2)×10−8
⇒13039.2×10−8Sm−1