Question

Which of the following is the correct expression for molar conductivity of \(\text{Na}_{2}\text{SO}_{4}\) at infinite dilution?

• A. \(\Lambda _{0}(\text{Na}_{2}\text{SO}_{4})=2\lambda _{(\text{Na}^{+})}^{0}+\lambda ^{0}(\text{SO}_{4}^{2-})\)
• B. \(\Lambda _{0}(\text{Na}_{2}\text{SO}_{4})=\lambda _{(\text{Na}^{+})}^{0}+2\lambda ^{0}(\text{SO}_{4}^{2-})\)
• C. \(\Lambda _{0}(\text{Na}_{2}\text{SO}_{4})=\lambda ^{0}(\text{Na}^{+})+\lambda ^{0}(\text{SO}_{4}^{2-})\)
• D. \(\Lambda _{0}(\text{Na}_{2}\text{SO}_{4})=(1/2)\lambda ^{0}(\text{Na}^{+})+\lambda ^{0}(\text{SO}_{4}^{2-})\)

Hint:

Apply Kohlrausch's law with the number of each ion in the formula.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Kohlrausch's law states that limiting molar conductivity is the sum of the limiting ionic conductivities, each multiplied by its number in the formula.

Step 2: Dissociation
\(\text{Na}_{2}\text{SO}_{4}\rightarrow 2\text{Na}^{+}+\text{SO}_{4}^{2-}\)

Step 3: Write the Expression
\(\Lambda_{0}(\text{Na}_{2}\text{SO}_{4})=2\lambda^{0}(\text{Na}^{+})+\lambda^{0}(\text{SO}_{4}^{2-})\)

Step 4: Compare the options
This matches option (A).

Final Answer:
The expression has twice the sodium ion conductivity plus the sulphate ion conductivity. \[ \boxed{\text{(A) }\Lambda_{0}(\text{Na}_{2}\text{SO}_{4})=2\lambda_{(\text{Na}^{+})}^{0}+\lambda^{0}(\text{SO}_{4}^{2-})} \]