Question

Calculate molar conductivity of $\mathrm{NH_4OH}$ at infinite dilution if molar conductivity of $\mathrm{Ba(OH)_2}$, $\mathrm{BaCl_2}$ and $\mathrm{NH_4Cl}$ at infinite dilution are $520$, $280$, $129\ \Omega^{-1}\,\mathrm{cm^2\,mol^{-1}}$ respectively.

• A. $249.0\ \Omega^{-1}\,\mathrm{cm^2\,mol^{-1}}$
• B. $498.0\ \Omega^{-1}\,\mathrm{cm^2\,mol^{-1}}$
• C. $125.0\ \Omega^{-1}\,\mathrm{cm^2\,mol^{-1}}$
• D. $369.0\ \Omega^{-1}\,\mathrm{cm^2\,mol^{-1}}$

Hint:

Chemistry Tip: Kohlrausch law helps find conductivity of weak electrolytes.

Answer 1

The Correct Option is A

Step 1: By Kohlrausch law, $$ \Lambda^\circ(\mathrm{NH_4OH})=\Lambda^\circ(\mathrm{NH_4Cl})+\frac{1}{2}\Lambda^\circ(\mathrm{Ba(OH)_2})-\frac{1}{2}\Lambda^\circ(\mathrm{BaCl_2}) $$

Step 2: Substitute values: $$ =129+\frac{520}{2}-\frac{280}{2} $$

Step 3: Calculate: $$ =129+260-140=249 $$ $$ \therefore \Lambda^\circ(\mathrm{NH_4OH})=249.0\ \Omega^{-1}\,\mathrm{cm^2\,mol^{-1}} $$