Question

Degree of dissociation of pure $H _{2} O$ is $1.9 \times 10^{-9}$ Molar ionic conductances of $H ^{+} \& OH ^{-}$ ions at infinite dilution are $200 \,S\, cm ^{2} mol ^{-1} \& 350\, S\, cm ^{2}$ $mol ^{-1}$ respectively. Molar conductance of water is:

• A. $3.8 \times 10^{-7} S \,cm ^{2} mol ^{-1}$
• B. $5.7 \times 10^{-7} S \,cm ^{2} mol ^{-1}$
• C. $9.5 \times 10^{-7} S\, cm ^{2} mol ^{-1}$
• D. $1.045 \times 10^{-6} S \,cm ^{2} mol ^{-1}$

Answer 1

Answer: (D)

$\alpha=\frac{\Lambda_{ m }^{ c }}{\Lambda_{ m }^{\infty}}$

$\Lambda_{ m }^{\infty}\left( H _{2} O \right)=\lambda_{ m }^{\infty}\left( H ^{+}\right)+\lambda_{ m }^{\infty}\left( OH ^{-}\right)$

$\Rightarrow 1.9 \times 10^{-9}=\frac{\Lambda_{ m }^{ c }}{550\, S\,cm ^{2} mol ^{-1}}$ $=200+350 \Rightarrow 550$

$\Lambda_{ m }^{ c }=550 \times 1.9 \times 10^{-9} S\,cm ^{2} mol ^{-1}$

$\quad=1.045 \times 10^{-6}\, S\,cm ^{2} mol ^{-1}$