Question

The conductivities at infinite dilution of $\text{N}{\text{H}}_{\text{4}}\text{Cl,NaOH}$ and $\text{NaCl}$ are 130, 218, 120 $\text{oh}{\text{m}}^{\text{- 1}}\text{c}{\text{m}}^{\text{2}}\text{e}{\text{q}}^{\text{- 1}}$. If equivalent conductance of N/100 solution of $\text{N}{\text{H}}_{\text{4}}\text{OH}$ is 10, then degree of dissociation of $\text{N}{\text{H}}_{\text{4}}\text{OH}$ at this dilution is:

• A. 0.005
• B. 0.043
• C. 0.01
• D. 0.02

Answer 1

Answer: (B)

0.043

$N{H}_{4}Cl+NaOH⟶N{H}_{4}OH+NaCl$

$\therefore {\wedge }_m^{\infty }$ ${}_{N{H}_{4}OH}={\wedge }_m^{\infty }$

${}_{N{H}_{4}Cl}+{\wedge }_m^{\infty }$ $NaOH-{\wedge }_m^{\infty }$ ${}_{NaCl}$

${\wedge }_m^{\infty }=130+218-120$

$⟹{\wedge }_m^{\infty }=228$ $sc{m}^{2}e{q}^2$

$\alpha =\frac{{\wedge }_m}{{\wedge }_{m^{\infty }}}=\frac{10}{228}=0.0438$