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 \(\text{S cm}^2\text{ mol}^{-1}\).
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 \(\text{S cm}^2\text{ mol}^{-1}\).
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Use $\Lambda_m=\frac{\kappa\times1000}{M}$ with $\kappa$ in S cm$^{-1}$.
Updated On: Oct 1, 2026
- \(125\)
- \(62.5\)
- \(12.5\)
- \(250\)
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The Correct Option is A
Solution and Explanation
Step 1: Write the formula
Molar conductivity links to specific conductance by \(\Lambda_m=\frac{\kappa\times1000}{M}\), where \(M\) is the molarity and the factor 1000 converts cm\(^3\) to litres.
Step 2: Substitute
\[ \Lambda_m=\frac{2.50\times10^{-4}\times1000}{0.0020}=\frac{0.25}{0.0020}=125\text{ S cm}^2\text{ mol}^{-1} \]
Step 3: Check the options
\(62.5\) comes from using \(0.004\) M, \(12.5\) from dropping a factor of 10, and \(250\) from using \(0.001\) M. The correct value is \(125\), option (A).
Final Answer:
\(\Lambda_m=125\) S cm\(^2\) mol\(^{-1}\), option (A). \[ \boxed{\text{(A) } 125} \]
Molar conductivity links to specific conductance by \(\Lambda_m=\frac{\kappa\times1000}{M}\), where \(M\) is the molarity and the factor 1000 converts cm\(^3\) to litres.
Step 2: Substitute
\[ \Lambda_m=\frac{2.50\times10^{-4}\times1000}{0.0020}=\frac{0.25}{0.0020}=125\text{ S cm}^2\text{ mol}^{-1} \]
Step 3: Check the options
\(62.5\) comes from using \(0.004\) M, \(12.5\) from dropping a factor of 10, and \(250\) from using \(0.001\) M. The correct value is \(125\), option (A).
Final Answer:
\(\Lambda_m=125\) S cm\(^2\) mol\(^{-1}\), option (A). \[ \boxed{\text{(A) } 125} \]
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