Question

At \(298\) K, the specific conductance (\(κ\)) of a \(0.0020\) M NaCl solution is \(2.50\times 10^{-4}\text{ S cm}^{-1}\). Calculate the molar conductivity (\(\Lambda _m\)) of the solution in \(\text{S cm}^2\text{ mol}^{-1}\).

• A. \(125\)
• B. \(62.5\)
• C. \(12.5\)
• D. \(250\)

Hint:

Use $\Lambda_m=\frac{\kappa\times1000}{M}$ with $\kappa$ in S cm$^{-1}$.

Answer 1

The Correct Option is A

Step 1: Convert the concentration
$0.0020$ mol/L is $0.0020\times10^{-3}=2.0\times10^{-6}$ mol/cm$^3$.

Step 2: Divide $\kappa$ by concentration
$\Lambda_m=\frac{\kappa}{c}=\frac{2.50\times10^{-4}}{2.0\times10^{-6}}=125$ S cm$^2$ mol$^{-1}$.
This is option (A).

Final Answer:
$\Lambda_m=125$, option (A). \[ \boxed{\text{(A)}} \]