JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
\(1 \times 10^{-5}\,M \textrm {AgNO } _ { 3 }\) is added to \(1\,L\) of saturated solution of \(AgBr\). The conductivity of this solution at \(298\,K\) is \(.........\times 10^{-8}\,S\,m ^{-1}\) \(\left[\right.\) Given : \(K _{ sp }( AgBr )=4.9 \times 10^{-13}\) at \(298 K\) \(\lambda_{ Ag ^{+}}^0=6 \times 10^{-3} Sm ^2\,mol ^{-1}\) \(\lambda_{ Br ^{-}}^0=8 \times 10^{-3} Sm ^2\,mol ^{-1}\) \(\left.\lambda_{ NO _3^{-}}^0=7 \times 10^{-3} Sm ^2\,mol ^{-1}\right]\)
- A \(12\)
- B \(14\)
- C \(13\)
- D \(15\)
Answer & Solution
Correct Answer
(B) \(14\)
Step-by-step Solution
Detailed explanation
\({\left[ Ag ^{+}\right]=10^{-5}}\) \({\left[ NO _3^{-}\right]=10^{-5}}\) \({\left[ Br ^{-}\right]=\frac{ Ksp }{\left[ Ag ^{+}\right]}=4.9 \times 10^{-8}}\) \(\Lambda_{ m }=\frac{ k }{1000 \times M }\) For \(Ag ^{+}\)…
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