CBSE 2025, Set 3

Question

$\Lambda^\circ_m$ NH$_4$Cl, $\Lambda^\circ_m$ NaOH and $\Lambda^\circ_m$ NaCl are 129.8, 217.4, and 108.9 S cm$^2$ mol$^{-1}$ respectively. Molar conductivity of $1 	imes 10^{-2}$ M solution of NH$_4$OH is 9.33 S cm$^2$ mol$^{-1}$. Calculate the degree of dissociation ($\alpha$) of NH$_4$OH solution at this concentration.

Edvaya · Accepted Answer

The correct answer is Subjective. **Step 1:** Calculate $\Lambda^\circ_m(	ext{NH}_4	ext{OH})$ using Kohlrausch's law: $$\Lambda^\circ_m(	ext{NH}_4	ext{OH}) = \Lambda^\circ_m(	ext{NH}_4	ext{Cl}) + \Lambda^\circ_m(	ext{NaOH}) - \Lambda^\circ_m(	ext{NaCl})$$ $$= 129.8 + 217.4 - 108.9 = \boxed{238.3 	ext{ S cm}^2 	ext{ mol}^{-1}}$$ **Step 2:** Calculate degree of dissociation $\alpha$: $$\alpha = \frac{\Lambda_m^c}{\Lambda^\circ_m} = \frac{9.33}{238.3} = \boxed{0.039 \approx 3.9\%}$$ The degree of dissociation of NH$_4$OH at $1 	imes 10^{-2}$ M is **0.039 (3.9%)**.

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