Question

The conductivity of 0.3 M solution of KCl at 298 K is 0.0627 S cm$^{-1}$. What is it's molar conductivity?

• A. 104 S cm$^2$ mol$^{-1}$
• B. 188 S cm$^2$ mol$^{-1}$
• C. 209 S cm$^2$ mol$^{-1}$
• D. 109 S cm$^2$ mol$^{-1}$

Hint:

Dividing by $0.3$ is exactly the same as multiplying the numerator by $\frac{10}{3}$. After multiplying $0.0627$ by $1000$ to get $62.7$, just think of it as roughly $\frac{63}{3} = 21$. This instant approximation leads straight to $209$.

Answer 1

The Correct Option is C

Step 1: Identify what to find.
We want molar conductivity $\Lambda_m$ from the conductivity $\kappa$ and the molarity $M$.
Step 2: Write the relation.
$\Lambda_m = \dfrac{\kappa \times 1000}{M}$, where the 1000 converts cm$^3$ to the litre used in molarity.
Step 3: List the data.
$\kappa = 0.0627$ S cm$^{-1}$ and $M = 0.3$ mol L$^{-1}$.
Step 4: Multiply by 1000.
$0.0627 \times 1000 = 62.7$.
Step 5: Divide by molarity.
$\Lambda_m = \dfrac{62.7}{0.3} = 209$.
Step 6: Attach units and choose.
$\Lambda_m = 209$ S cm$^2$ mol$^{-1}$, which is option (3).
\[ \boxed{\Lambda_m = 209 \text{ S cm}^2 \text{ mol}^{-1}} \]