Question

The molar conductivity of 0.01 M acetic acid at 25 °C is 16.5 $\Omega^{-1}$ cm² mol$^{-1}$ and its molar conductivity at zero concentration is 390.7 $\Omega^{-1}$ cm² mol$^{-1}$. What is its degree of dissociation?

• A. 0.0223
• B. 0.0422
• C. 0.0642
• D. 0.0821

Answer 1

Answer: (B)

0.0422

Answer 2

The degree of dissociation $\alpha$ is given by:

$$\alpha = \frac{\Lambda}{\Lambda^0}$$

Where:

• $\Lambda = 16.5\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$
• $\Lambda^0 = 390.7\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$

Thus,

$$\alpha = \frac{16.5}{390.7} \approx 0.0422$$

Degree of dissociation is the ratio of the given and limiting molar conductivities, i.e., $\alpha = \frac{16.5}{390.7} \approx 0.0422$.