Question

What is percent dissociation of $NH_4OH$ if molar conductance at zero concentration for $NH_4Cl, NaCl$ and $NaOH$ are 130, 109 and 213 S cm$^2$ mol$^{-1}$ respectively and molar conductivity of 0.01 M $NH_4OH$ is 9.0 S cm$^2$ mol$^{-1}$ ?

• A. $\frac{100}{40}$
• B. $\frac{100}{35}$
• C. $\frac{100}{32}$
• D. $\frac{100}{26}$

Hint:

$\alpha = \frac{\Lambda_m}{\Lambda^\circ_m}$. Use Kohlrausch's Law to find $\Lambda^\circ_m$ for weak electrolytes.

Answer 1

The Correct Option is D

Step 1: Concept
For a weak electrolyte, degree of dissociation ($\alpha$) is given by: \[ \alpha = \frac{\Lambda_m}{\Lambda_m^\circ} \] where $\Lambda_m^\circ$ is the molar conductivity at infinite dilution. Using Kohlrausch’s law: \[ \Lambda_m^\circ (NH_4OH) = \Lambda_m^\circ (NH_4Cl) + \Lambda_m^\circ (NaOH) - \Lambda_m^\circ (NaCl) \]

Step 2: Calculation of $\Lambda_m^\circ (NH_4OH)$
\[ \Lambda_m^\circ (NH_4OH) = 130 + 213 - 109 = 234 \, \text{S cm}^2 \text{ mol}^{-1} \]

Step 3: Degree of dissociation
Given: \[ \Lambda_m = 9.0 \, \text{S cm}^2 \text{ mol}^{-1} \] \[ \alpha = \frac{9}{234} = \frac{1}{26} \]

Step 4: Percent dissociation
\[ % \text{ dissociation} = \alpha \times 100 = \frac{100}{26} \] {red{
Final Answer: (D)