Chemistry - Electrochemistry Question with Solution | TestHub

${\mathrm{K}}_{{\mathrm{H}}_2\mathrm{O}}=\frac{C.C}{9000}$

For KCl solution

$({\Lambda }_{\mathrm{M}}{)}_{\mathrm{KCl}}=\frac{{\mathrm{K}}_{\mathrm{KCl}}\times 1000}{0.02}=150$

${\mathrm{K}}_{\mathrm{KCl}}=300\times 10^{-5}{\Omega }^{-1}{\mathrm{cm}}^{-1}(\mathrm{A})$ is correct ${\mathrm{K}}_{\mathrm{KCl}\mathrm{soln}}=\frac{C.C}{90}={\mathrm{K}}_{\mathrm{KCl}}+{\mathrm{K}}_{{\mathrm{H}}_2\mathrm{O}}$

$$\begin{array}{rl}\frac{C.C}{90}& =300\times 10^{-5}+\frac{C.C}{9000}\\ C.C\left[\frac{1}{90}-\frac{1}{9000}\right]& =300\times 10^{-5}\\ C.C\left[\frac{100-1}{9000}\right]& =300\times 10^{-5}\\ C.C\left[\frac{99}{9000}\right]& =300\times 10^{-5}\\ C.C& =\frac{300\times 10^{-5}\times 9000}{99}\\ C.C& =\frac{2700000\times 10^{-5}}{99}\\ C.C& =\frac{27}{99}=\frac{3}{11}{\mathrm{cm}}^{-1}\\ {\mathrm{K}}_{{\mathrm{H}}_2\mathrm{O}}& =\frac{3}{11\times 9000}=\frac{1}{33000}=\frac{1}{33}\times 10^{-3}\\ {\mathrm{K}}_{{\mathrm{H}}_2\mathrm{O}}& =\frac{100}{33}\times 10^{-5}\\ {\mathrm{K}}_{\mathrm{NaCl}\mathrm{soln}}& ={\mathrm{K}}_{\mathrm{NaCl}}+{\mathrm{K}}_{{\mathrm{H}}_2\mathrm{O}}\\ ({\Lambda }_{\mathrm{M}}{)}_{\mathrm{NaCl}}& =\frac{{\mathrm{K}}_{\mathrm{NaCl}}\times 1000}{[\mathrm{NaCl}]}\\ {\mathrm{K}}_{\mathrm{NaCl}}& =120\times 10^{-3}[\mathrm{NaCl}]\\ {\mathrm{K}}_{\mathrm{NaCl}\mathrm{soln}}& =\frac{3}{11\times 7000}=\frac{3}{77000}=\frac{3}{77}\times 10^{-3}=\frac{300}{77}\times 10^{-5}{\Omega }^{-1}{\mathrm{cm}}^{-1}\\ {\mathrm{K}}_{\mathrm{NaCl}}& =\left(\frac{300}{77}-\frac{100}{33}\right)\times 10^{-5}\\ 120\times 10^{-3}[\mathrm{NaCl}]& =\left(\frac{9900-7700}{77\times 33}\right)\times 10^{-5}\\ 120\times 10^{-3}[\mathrm{NaCl}]& =\frac{2200}{2541}\times 10^{-5}\\ [\mathrm{NaCl}]& =\frac{2200}{2541\times 120}\times 10^{-2}\\ [\mathrm{NaCl}]& =\frac{22}{2541\times 12}\times 10^{-2}\\ [\mathrm{NaCl}]& =\frac{11}{2541\times 6}\times 10^{-2}\\ [\mathrm{NaCl}]& =\frac{11}{15246}\times 10^{-2}\\ \frac{10}{\mathrm{V}(\ell \mathrm{t})}& =\frac{11}{15246}\times 10^{-2}\\ \mathrm{V}(\ell \mathrm{t})& =\frac{10\times 15246}{11\times 10^{-2}}=\frac{152460}{11}\times 10^2\\ \mathrm{V}(\ell \mathrm{t})& =13860\times 10^2=1386000\ell \mathrm{t}\end{array}$$