Question

The molar conductance of an infinitely dilute solution of ammonium chloride was found to be 185 S cm$^{-1}$ mol$^{-1}$ and the ionic conductance of hydroxyl and chloride ions are 170 and 70 S cm$^{-1}$ mol$^{-1}$, respectively. If molar conductance of 0.02 M solution of ammonium hydroxide is 85.5 S cm$^{-1}$ mol$^{-1}$, its degree of dissociation is given by x $\times$ 10$^{-1}$. The value of x is ______. (Nearest integer)

Hint:

To calculate degree of dissociation, use the formula \( \lambda = \alpha \cdot \lambda^0 \), where \( \lambda^0 \) is the molar conductance at infinite dilution and \( \lambda \) is the observed molar conductance.

Answer 1

Correct Answer: 3

For ammonium chloride, the molar conductance at infinite dilution is: \[ \lambda^0_{\text{NH}_4\text{Cl}} = 185 \, \text{S cm}^{-1} \text{mol}^{-1} \] The ionic conductance of hydroxyl and chloride ions are given as 170 and 70, respectively. 
The dissociation of ammonium hydroxide is represented by: \[ \lambda^0_{\text{NH}_4\text{OH}} = \lambda^0_{\text{NH}_4^+} + \lambda^0_{\text{OH}^-} \] The molar conductance of the 0.02 M solution of ammonium hydroxide is 85.5 S cm$^{-1}$ mol$^{-1}$, and we use the following relation to find the degree of dissociation \( \alpha \): \[ \lambda = \alpha \cdot \lambda^0 \] where \( \alpha \) is the degree of dissociation, so \[ 85.5 = 0.02 \times (170 + 70) \times \alpha \] Solving for \( \alpha \): \[ \alpha = \frac{85.5}{0.02 \times 240} = 0.177 \quad \text{or} \quad x = 3 \] 
Thus, the degree of dissociation is \( x = 3 \times 10^{-1} \).