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: Plan
Find the number of moles in one cm$^3$ first, then convert to litres.

Step 2: Moles per cm3
$\kappa = \Lambda_m \times (\text{mol/cm}^3)$, so mol per cm$^3$ = $0.012/240 = 5 \times 10^{-5}$.

Step 3: Convert
Multiply by 1000 cm$^3$ per litre: $5 \times 10^{-5} \times 1000 = 0.05$ mol/L.

Step 4: Answer
The molarity is 0.05 M, option (C).

Final Answer:
Molarity of BaCl2 is 0.05 M. This is option (C). \[ \boxed{\text{(C) }0.05\ \text{M}} \]