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: Understanding the Question:
We need to calculate the percent dissociation ($\alpha \times 100$) of the weak base $NH_4OH$. This requires calculating its molar conductivity at infinite dilution ($\Lambda_m^\circ$) using Kohlrausch's Law.
Step 2: Key Formula or Approach:
1. Kohlrausch's Law: $\Lambda_m^\circ (NH_4OH) = \Lambda_m^\circ (NH_4Cl) + \Lambda_m^\circ (NaOH) - \Lambda_m^\circ (NaCl)$
2. Degree of dissociation: $\alpha = \frac{\Lambda_m^c}{\Lambda_m^\circ}$
3. Percent dissociation: $% \alpha = \alpha \times 100$
Step 3: Detailed Explanation:
Given:
$\Lambda_m^\circ (NH_4Cl) = 130$
$\Lambda_m^\circ (NaOH) = 213$
$\Lambda_m^\circ (NaCl) = 109$
$\Lambda_m^c (NH_4OH) = 9.0$
Calculating $\Lambda_m^\circ$ for $NH_4OH$:
\[ \Lambda_m^\circ (NH_4OH) = 130 + 213 - 109 = 343 - 109 = 234\text{ S cm}^2\text{ mol}^{-1} \]
Calculating $\alpha$:
\[ \alpha = \frac{9.0}{234} = \frac{1}{26} \]
Calculating percent dissociation:
\[ % \text{ dissociation} = \frac{1}{26} \times 100 = \frac{100}{26} % \]
Step 4: Final Answer:
The percent dissociation is $\frac{100}{26}$.