Question

Following limiting molar conductivities are given as:

\[\ce{\lambda^{\circ}_{m} (H2SO4)}\] = x S cm² mol⁻¹

\[\ce{\lambda{^{\circ}_{m}} (K2SO4)}\] = y S cm² mol⁻¹

\[\ce{\lambda{^{\circ}_{m}} (CH3COOK)}\] = z S cm² mol⁻¹

\[\ce{\lambda^{\circ}_{m}}\] (in S cm² mol⁻¹) for CH₃COOH will be:

Options

• \[\ce{\frac{(x - y)}{2} + z}\]

• x − y + 2z

• x + y + z

• x − y + z

Answer 1

\[\ce{\frac{(x - y)}{2} + z}\]

Explanation:

Using Kohlrausch’s Law of Independent Ionic Migration, we can find the limiting molar conductivity of acetic acid:

\[\ce{\lambda^{\circ}_m({CH_3COOH}) = \lambda^{\circ}_m({CH_3COOK}) + \frac{1}{2}[\lambda^{\circ}_m({H_2SO_4}) - \lambda^{\circ}_m({K_2SO_4})}\]

= \[\ce{x + \frac{(x - y)}{2}}\]

= \[\ce{\frac{(x - y)}{2} + x}\]