JEE Main 2023 — Chemistry Physical Chemistry

The concentrations of the ions can be calculated as follows,

$$[{\mathrm{Ag}}^{+}]={10}^{-5}M$$[Ag+]=10−5M

$$[{\mathrm{NO}}_{3}^{-}]={10}^{-5}M$$[NO3−​]=10−5M

$$[{\mathrm{Br}}^{-}]=\frac{\mathrm{Ksp}}{[{\mathrm{Ag}}^{+}]}=4.9\times {10}^{-8}M$$[Br−]=[Ag+]Ksp​=4.9×10−8M

$${\Lambda }_{m}=\frac{k}{1000\times M}=\frac{\mathrm{Specific}\mathrm{conductance}}{1000\times \mathrm{Molarity}}$$Λm​=1000×Mk​=1000×MolaritySpecificconductance​

For ${\mathrm{Ag}}^{+}$Ag+

$$6\times {10}^{-3}=\frac{{k}_{{\mathrm{Ag}}^{+}}}{1000\times {10}^{-5}}$$6×10−3=1000×10−5kAg+​​

$${k}_{\mathrm{Ag}+}=6\times {10}^{-5}{\mathrm{Sm}}^{-1}$$kAg+​=6×10−5Sm−1

$$\Rightarrow 6000\times {10}^{-8}{\mathrm{Sm}}^{-1}$$⇒6000×10−8Sm−1

for ${\mathrm{Br}}^{-}$Br−

$$8\times {10}^{-3}=\frac{{k}_{{\mathrm{Br}}^{-}}}{1000\times 4.9\times {10}^{-8}}$$8×10−3=1000×4.9×10−8kBr−​​

$${k}_{\mathrm{Br}-}=39.2\times {10}^{-8}{\mathrm{Sm}}^{-1}$$kBr−​=39.2×10−8Sm−1

for ${\mathrm{NO}}_{3}^{-}$NO3−​

$$7\times {10}^{-3}=\frac{{k}_{{\mathrm{NO}}_{3}^{-}}}{1000\times {10}^{-5}}$$7×10−3=1000×10−5kNO3−​​​

$${k}_{{\mathrm{NO}}_{3}^{-}}=7\times {10}^{-5}{\mathrm{Sm}}^{-1}$$kNO3−​​=7×10−5Sm−1

$$=7000\times {10}^{-8}{\mathrm{Sm}}^{-1}$$=7000×10−8Sm−1

Conductivity of solution

$$\Rightarrow (6000+7000+39.2)\times {10}^{-8}$$⇒(6000+7000+39.2)×10−8

$$\Rightarrow 13039.2\times {10}^{-8}S{m}^{-1}$$⇒13039.2×10−8Sm−1