Question

Specific conductance of 0.1 M HNO\(_3\) is \(6.3 \times 10^{-2} \, \text{ohm}^{-1} \, \text{cm}^{-1}\). The molar conductance of the solution is:

• A. \(100 \, \text{ohm}^{-1} \, \text{cm}^{2} \, \text{mol}^{-1}\)
• B. \(515 \, \text{ohm}^{-1} \, \text{cm}^{2} \, \text{mol}^{-1}\)
• C. \(630 \, \text{ohm}^{-1} \, \text{cm}^{2} \, \text{mol}^{-1}\)
• D. \(6300 \, \text{ohm}^{-1} \, \text{cm}^{2} \, \text{mol}^{-1}\)

Hint:

Molar conductance (\( \Lambda_m \)) is calculated using the formula: \[ \Lambda_m = \frac{\kappa \times 1000}{C}, \] where \( \kappa \) is the specific conductance and \( C \) is the concentration in mol/L.

Answer 1

The Correct Option is C

Step 1: Understanding the Terms
Specific Conductance (\( \kappa \)): Given as \(6.3 \times 10^{-2} \, \text{ohm}^{-1} \, \text{cm}^{-1}\).
Molar Conductance (\( \Lambda_m \)): This is the conductance of all ions produced by 1 mole of the electrolyte in a solution.
Step 2: Formula for Molar Conductance
The molar conductance (\( \Lambda_m \)) is calculated using the formula: \[ \Lambda_m = \frac{\kappa \times 1000}{C}, \] where:
\( \kappa \) is the specific conductance,
\( C \) is the concentration of the solution in mol/L (0.1 M in this case).
Step 3: Substituting the Values
Substitute the given values into the formula: \[ \Lambda_m = \frac{6.3 \times 10^{-2} \times 1000}{0.1}. \] \[ \Lambda_m = \frac{63}{0.1} = 630 \, \text{ohm}^{-1} \, \text{cm}^{2} \, \text{mol}^{-1}. \]
Step 4: Matching with the Options
The calculated molar conductance is \(630 \, \text{ohm}^{-1} \, \text{cm}^{2} \, \text{mol}^{-1}\), which corresponds to option (C). Final Answer: The molar conductance of the solution is (C) \(630 \, \text{ohm}^{-1} \, \text{cm}^{2} \, \text{mol}^{-1}\).