Question

If ${\wedge }_{c}ofN{H}_{4}OHis11.5{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$ , its degree of dissociation would be:

(Given ${\lambda }_{N{H}_4^{+}}^{\infty }=73.4{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$ and ${\lambda }_{O{H}^{-}}^{\infty }=197.6{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$) :

• A. $0.157$
• B. $0.058$
• C. $0.0424$
• D. $0.0848$

Answer 1

The correct option is C $0.0424$

${\Lambda }_{N{H}_{4}OH}^{\infty }={\lambda }_{N{H}_4^{+}}^{\infty }+{\lambda }_{O{H}^{-}}^{\infty }$

Substitute values in the above equation:

${\Lambda }_{N{H}_{4}OH}^{\infty }=73.4+197.6=271{\Omega }^{-1}{cm}^2{mol}^{-1}$

The degree of dissociation is the ratio of the equivalent conductance at given concentration to the equivalent conductance at zero concentration. Hence,

$\alpha =\frac{{\Lambda }_c}{{\Lambda }_{\infty }}=\frac{11.5}{271}=\mathrm{0.0424.}$