Question

Calculate the concentration of silver nitrate solution if molar conductivity and conductivity of silver nitrate at \(25^{\circ}\)C are respectively 120 ohm\(^{-1}\) cm\(^2\) mol\(^{-1}\) and 0.0024 ohm\(^{-1}\) cm\(^{-1}\).

• A. \(0.01\) M
• B. \(0.02\) M
• C. \(0.03\) M
• D. \(0.04\) M

Hint:

Molar conductivity = (conductivity x 1000) / concentration.

Answer 1

The Correct Option is B

Step 1: Relation
$\Lambda_m = 1000\kappa / c$ with $c$ in mol/L.

Step 2: Rearrange
$c = 1000 \times 0.0024 / 120$.

Step 3: Evaluate
$c = 2.4/120 = 0.02$ M, option (B).

Final Answer:
Option (B). \[ \boxed{0.02 \text{ M}} \]