Question

The molar conductivity of \(0.01\) M monobasic acid at \(25\,^{\circ}\text{C}\) is \(15\) ohm\(^{-1}\) cm\(^2\) mol\(^{-1}\). The molar conductivity of the same acid at infinite dilution is \(375\) ohm\(^{-1}\) cm\(^2\) mol\(^{-1}\). Calculate \([\text{H}^+]\) in solution.

• A. \(2.0\times 10^{-4}\) M
• B. \(3.0\times 10^{-4}\) M
• C. \(4.0\times 10^{-4}\) M
• D. \(5.0\times 10^{-4}\) M

Hint:

alpha = molar conductivity / limiting molar conductivity.

Answer 1

The Correct Option is C

Step 1: Fraction ionised:
Only a part of the acid has ionised. The molar conductivity is $15$ against a possible $375$ for full ionisation, so the fraction is $15/375 = 1/25$.

Step 2: Apply to concentration:
One acid molecule gives one $H^+$. So $[H^+] = 0.01\times (1/25) = 4\times 10^{-4}$ M.

Step 3: Match:
Option (C).

Final Answer:
1/25 of the 0.01 M acid is ionised. \[ \boxed{C} \]