Question

What will be the concentration of solution of electrolyte if its molar conductivity and conductivity are respectively 230 \(\Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1}\) and 0.0115 \(\Omega^{-1} \, \text{cm}^{-1}\) at 298 K?

• A. 0.04 M
• B. 0.03 M
• C. 0.01 M
• D. 0.05 M

Hint:

Always ensure units are consistent: \(\kappa\) in S/cm, \(\Lambda_m\) in S cm²/mol gives \(C\) in mol/L. The factor 1000 converts cm³ to L.

Answer 1

The Correct Option is D

Step 1: Understand the question.
We know molar conductivity $\Lambda_m = 230\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}$ and conductivity $\kappa = 0.0115\ \Omega^{-1}\text{cm}^{-1}$. We need the concentration $C$.
Step 2: Recall the linking formula.
The three quantities are tied together by: \[ \Lambda_m = \frac{\kappa \times 1000}{C} \] where $C$ is in mol per litre.
Step 3: Rearrange for C.
Swap $C$ and $\Lambda_m$: \[ C = \frac{\kappa \times 1000}{\Lambda_m} \]
Step 4: Put the numbers in.
\[ C = \frac{0.0115 \times 1000}{230} \]
Step 5: Do the arithmetic.
The top is $0.0115 \times 1000 = 11.5$. Then: \[ C = \frac{11.5}{230} = 0.05 \text{ mol/L} \]
Step 6: Choose the answer.
The concentration is 0.05 M, which is option 4. \[ \boxed{0.05\ \text{M}} \]