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

To determine the degree of dissociation (α) of ammonium hydroxide (NH₄OH), we use the following relationship for molar conductance:

λ_m = λ₀ α

Where:

• λ_m = molar conductance of the solution (85.5 S cm^-1 mol^-1)
• λ₀ = molar conductance at infinite dilution of NH₄OH
• α = degree of dissociation

First, calculate λ₀ using the ion conductances:

λ₀(NH₄OH) = λ₀(NH₄Cl) - λ_m(Cl^-) + λ_m(OH^-)

= 185 S cm^-1 mol^-1 - 70 S cm^-1 mol^-1 + 170 S cm^-1 mol^-1

= 285 S cm^-1 mol^-1

Now, substitute the values into the degree of dissociation formula:

α = λ_m / λ₀

= 85.5 / 285

= 0.3

To express α as x×10^-1, we have x = 3.

Confirming against the expected range (3,3), the computed value x = 3 is within the range.

Therefore, the value of x is 3.

Answer 2

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 \( x = 3 \times 10^{-1} \).