Question

A $0.5MNaOH$ solution offers a resistance of $31.6ohm$ in a conductivity cell at room temperature. What shall be the approximate molar conductance of this $NaOH$ solution if cell constant of the cell is $0.367c{m}^{-1}$?

• A. $234Sc{m}^{2}mol{e}^{-1}$
• B. $23.2Sc{m}^{2}mol{e}^{-1}$
• C. $4645Sc{m}^{2}mol{e}^{-1}$
• D. $5464Sc{m}^{2}mol{e}^{-1}$

Answer 1

Answer: (B)

$23.2Sc{m}^{2}mol{e}^{-1}$

Here, $R=31.6ohm$

$\therefore G=\frac{1}{R}=\frac{1}{31.6}oh{m}^{-1}=0.0316oh{m}^{-1}$

Specific conductance (k)

$k=conductance\times$ cell constant

$k=0.0316oh{m}^{-1}\times 0.367c{m}^{-1}$

$k=0.0116oh{m}^{-1}c{m}^{-1}$

Now, molar concentration $=0.5M\left(given\right)$

$=0.5\times {10}^{-3}molec{m}^{-3}$

$\therefore$ Molar conductance $=\frac{k}{\text{molar conc.}}$

${\Lambda }_m=\frac{0.0116}{0.5\times {10}^{-3}}$

${\Lambda }_m=23.2Sc{m}^{2}mo{l}^{-1}$

Hence, option B is the correct option.