The molar conductance of an infinitely dilute solution of ammonium chloride was found to be 185 S cm$$^2$$ mol$$^{-1}$$ and the ionic conductance of hydroxyl and chloride ions are 170 and 70 S cm$$^2$$ mol$$^{-1}$$, respectively. If molar conductance of 0.02 M solution of ammonium hydroxide is 85.5 S cm$$^2$$ mol$$^{-1}$$, its degree of dissociation is given by $$x \times 10^{-1}$$. The value of x is ______. (Nearest integer)

For weak electrolytes such as $$NH_4OH$$ the degree of dissociation $$\alpha$$ is obtained from
$$\alpha \;=\;\dfrac{\Lambda_m(c)}{\Lambda_m^{\infty}}$$
where $$\Lambda_m(c)$$ is the molar conductance at concentration $$c$$ and $$\Lambda_m^{\infty}$$ is the molar conductance at infinite dilution.

Step 1: Calculate the ionic conductance of the ammonium ion.

By Kohlrausch’s Law of Independent Migration of Ions, for any electrolyte
$$\Lambda_m^{\infty} \;=\; \lambda_{+}^{\infty} \;+\; \lambda_{-}^{\infty}$$

For $$NH_4Cl$$,
$$\Lambda_m^{\infty}(NH_4Cl) \;=\; 185\;S\;cm^{2}\;mol^{-1}$$
and the ionic conductance of $$Cl^-$$ is
$$\lambda_{Cl^-}^{\infty} \;=\; 70\;S\;cm^{2}\;mol^{-1}$$

Hence,
$$\lambda_{NH_4^+}^{\infty} \;=\; \Lambda_m^{\infty}(NH_4Cl)\;-\;\lambda_{Cl^-}^{\infty}$$ $$\lambda_{NH_4^+}^{\infty} \;=\; 185\;-\;70 \;=\; 115\;S\;cm^{2}\;mol^{-1}$$

Step 2: Obtain $$\Lambda_m^{\infty}$$ for $$NH_4OH$$.

For $$NH_4OH$$ the ions present are $$NH_4^+$$ and $$OH^-$$, so
$$\Lambda_m^{\infty}(NH_4OH) \;=\; \lambda_{NH_4^+}^{\infty} \;+\; \lambda_{OH^-}^{\infty}$$

The ionic conductance of $$OH^-$$ is
$$\lambda_{OH^-}^{\infty} \;=\; 170\;S\;cm^{2}\;mol^{-1}$$

Therefore,
$$\Lambda_m^{\infty}(NH_4OH) \;=\; 115 \;+\; 170 \;=\; 285\;S\;cm^{2}\;mol^{-1}$$

Step 3: Compute the degree of dissociation.

Given molar conductance of $$0.02\;M$$ $$NH_4OH$$ solution:
$$\Lambda_m(0.02M) \;=\; 85.5\;S\;cm^{2}\;mol^{-1}$$

Hence,
$$\alpha \;=\;\dfrac{\Lambda_m(0.02M)}{\Lambda_m^{\infty}(NH_4OH)} \;=\;\dfrac{85.5}{285}$$

$$\alpha \;=\; 0.30 \;=\; 3.0 \times 10^{-1}$$

Step 4: Expressing in the required form.

The question states $$\alpha = x \times 10^{-1}$$. Comparison gives $$x \approx 3$$ (nearest integer).

Hence, the required value of $$x$$ is 3.