Question

A conductivity cell, when filled with an aqueous solution of $0.02M$ $KCl$ at $25°C$, had a resistance of $250\text{ohm}$. Its resistance, when filled with $6\times {10}^{-5}MN{H}_{4}OH$ solution was ${10}^5$ $\text{ohm}$. The specific conducatnce of $0.02M$ $KCl$ was $0.277S{m}^{-1}$. The molar ionic conductances at infinite dilution of $N{H}_4^{+}\text{and}O{H}^{-}\text{ions are}73.4\times {10}^{-4}\text{and}198.0\times {10}^{-4}S{m}^{2}mo{l}^{-1},$ respectively. Calculate the degree of dissociation of $6\times {10}^{-5}MN{H}_{4}OH$ solution:

• A. 0.424
• B. 0.845
• C. 0.954
• D. 0.765

Answer 1

The correct option is A 0.424

Since specific conductance $\left(\kappa \right)=$ cell constant/R, hence

cell constant $=\kappa R=\left(0.277S{m}^{-1}\right)\left(250\Omega \right)=69.2m^{-1}(S={\Omega }^{-1})$

For $N{H}_{4}OH$ solution, $c=6\times {10}^{-5}M=6\times {10}^{-5}mold{m}^{-3}=6\times {10}^{-2}mol{m}^3$

${\Lambda }_m=\frac{\kappa }{c}=\frac{\text{cell constant}}{c}=\frac{69.2m^{-1}}{(6\times {10}^{-2}mol{m}^{-3}\times {10}^5\Omega )}$

$=115\times {10}^{-4}S{m}^{2}mo{l}^{-1}$

According to Kohlrauch's law,

${\Lambda }_m^{\circ }={\lambda }_{+}^{\circ }+{\lambda }_{-}^{\circ }=(73.4+198.0)\times {10}^{-4}S{m}^{2}mo{l}^{-1}$

$=271.4\times {10}^{-4}S{m}^{2}mo{l}^{-1}$

$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }}=\frac{115\times {10}^{-4}S{m}^{2}mo{l}^{-1}}{271.4\times {10}^{-4}S{m}^{2}mo{l}^{-1}}=0.424$