Question

The conductivity of $0.012\ \text{M}$ NaBr solution is $2.67 \times 10^{-4}\ \text{S\ cm}^{-1}$. What is it's molar conductivity?

• A. $26.7\ \text{S\ cm}^2\ \text{mol}^{-1}$
• B. $32.04\ \text{S\ cm}^2\ \text{mol}^{-1}$
• C. $12.2\ \text{S\ cm}^2\ \text{mol}^{-1}$
• D. $22.2\ \text{S\ cm}^2\ \text{mol}^{-1}$

Hint:

When evaluating fractions with small decimals like $\frac{0.267}{0.012}$, shifting the decimal point completely to convert it into whole numbers ($\frac{267}{12}$) makes manual long division drastically faster and minimizes basic arithmetic slip-ups!

Answer 1

The Correct Option is D

Step 1: Distinguish the two quantities.
Conductivity $\kappa$ is per unit volume, while molar conductivity $\Lambda_m$ scales that up to one mole of electrolyte. We must convert from one to the other.
Step 2: Recall the bridging formula.
When $\kappa$ is in S cm$^{-1}$ and concentration $C$ is in mol L$^{-1}$, the relation is $\Lambda_m = \dfrac{1000\,\kappa}{C}$. The factor 1000 converts litres to cm$^3$.
Step 3: Note the given values.
$\kappa = 2.67 \times 10^{-4}$ S cm$^{-1}$ and $C = 0.012$ mol L$^{-1}$.
Step 4: Build the numerator.
$1000 \times 2.67 \times 10^{-4} = 2.67 \times 10^{-1} = 0.267$.
Step 5: Divide by the concentration.
$\Lambda_m = \dfrac{0.267}{0.012}$. Multiplying top and bottom by 1000 clears the decimals: $\dfrac{267}{12} = 22.25$.
Step 6: State the rounded answer.
Rounding to one decimal place gives $22.2$ S cm$^2$ mol$^{-1}$, which is option (4).
\[ \boxed{\Lambda_m \approx 22.2\ \text{S cm}^2\ \text{mol}^{-1}} \]