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: Per cm3 to per litre:
1 L = 1000 cm$^3$. A 0.02 M solution has 0.02 mol in 1000 cm$^3$, so 1 cm$^3$ holds $0.02/1000 = 2\times10^{-5}$ mol.

Step 2: Divide conductivity by mol per cm3:
\[ \Lambda_m = \frac{4.37\times10^{-4}}{2\times10^{-5}} = 21.85 \]

Step 3: Units:
$\Omega^{-1}\text{cm}^{-1} / (\text{mol cm}^{-3}) = \Omega^{-1}\text{cm}^{2}\text{mol}^{-1}$, as in the options.

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