Question

What must be the molarity of \(\text{BaCl}_2\) solution to have molar conductivity \(240\,\Omega ^{-1}\text{cm}^2\text{mol}^{-1}\) and conductivity \(0.012\,\Omega ^{-1}\text{cm}^{-1}\) at \(25^{\circ}\text{C}\) ?

• A. \(0.01\) M
• B. \(0.02\) M
• C. \(0.05\) M
• D. \(0.1\) M

Hint:

Use molar conductivity = 1000 x conductivity / molarity.

Answer 1

The Correct Option is C

Step 1: Understand the relation
Molar conductivity and conductivity are linked by \(\Lambda_m = \dfrac{\kappa \times 1000}{C}\), where \(C\) is the molarity in mol per litre and \(\kappa\) is in \(\Omega^{-1}\text{cm}^{-1}\).

Step 2: Rearrange
\[ C = \frac{\kappa \times 1000}{\Lambda_m} \]

Step 3: Substitute
\[ C = \frac{0.012 \times 1000}{240} = \frac{12}{240} = 0.05\ \text{M} \]

Step 4: Check the options
If the 1000 factor is forgotten we get \(5 \times 10^{-5}\), not listed. The values 0.01, 0.02 and 0.1 M would give \(\kappa\) = 0.0024, 0.0048 and 0.024, so they do not match.

Final Answer:
The molarity is 0.05 M. This is option (C). \[ \boxed{\text{(C) }0.05\ \text{M}} \]