Question

What is the molar conductivity of \(0.02\) M KI solution if its conductivity is \(4.37\times 10^{-4}\,\Omega ^{-1}\,\text{cm}^{-1}\) ?

• A. \(8.74 \Omega ^{-1}\,\text{cm}^2\text{mol}^{-1}\)
• B. \(21.85 \Omega ^{-1}\,\text{cm}^2\text{mol}^{-1}\)
• C. \(43.70 \Omega ^{-1}\,\text{cm}^2\text{mol}^{-1}\)
• D. \(13.65 \Omega ^{-1}\,\text{cm}^2\text{mol}^{-1}\)

Hint:

Molar conductivity = kappa x 1000 / molarity.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Molar conductivity \(\Lambda_m\) is the conductivity of all the ions produced by one mole of electrolyte in a solution volume that contains that one mole.

Step 2: Key Formula:
\[ \Lambda_m = \frac{\kappa \times 1000}{M} \]
where \(\kappa\) is in \(\Omega^{-1}\text{cm}^{-1}\) and \(M\) is in mol L\(^{-1}\); the factor 1000 converts litres to cm\(^3\).

Step 3: Detailed Explanation:
\[ \Lambda_m = \frac{4.37 \times 10^{-4} \times 1000}{0.02} = \frac{0.437}{0.02} = 21.85\ \Omega^{-1}\text{cm}^2\text{mol}^{-1} \]

Step 4: Why the other options are wrong.
8.74 is obtained by using 0.05 M instead of 0.02 M. 43.70 comes from dividing by 0.01 M, a factor-of-two slip. 13.65 does not follow from the given data.

Final Answer:
The molar conductivity is \(21.85\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}\), option (B). \[ \boxed{21.85\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}} \]