Question

If the values of $\Lambda_{\infty}$ of $NH _{4} Cl , NaOH$ and $NaCl$ are $130,217 $ and $109\, ohm ^{-1} \,cm ^{2}$ equiv $^{-1}$ respectively, the $\Lambda_{\infty}$ of $NH _{4} OH$ in $ohm ^{-1}\, cm ^{2}$ equiv $^{-1}$ is

• A. 238
• B. 196
• C. 22
• D. 456

Answer 1

Answer: (A)

$\Lambda_{\infty} NH _{4} Cl =\Lambda_{\infty} NH _{4}^{+}+\Lambda_{\infty} Cl ^{-} =130\,\,\,...(i)$

$\Lambda_{\infty} NaOH =\Lambda_{\infty} Na ^{+}+\Lambda_{\infty} OH ^{-} =217 \,\,\,...(ii)$

$\Lambda_{\infty} NaCl =\Lambda_{\infty} Na ^{+}+\Lambda_{\infty} Cl ^{-} =109\,\,\,...(iii)$

On adding Eq. (i) and (ii) and subtracting the Eq. (iii)

$\Lambda_{\infty} NH _{4}^{+}+\Lambda_{\infty} OH ^{-} =130+217-109 $

$ \Lambda_{\infty} NH _{4} OH =347-109 $

$=238 \,ohm ^{-1} \,cm ^{2} $ equiv $^{-1} $