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: Understanding the Question:
The question provides the concentration (molarity) and specific conductivity ($\kappa$) of a sodium bromide (NaBr) solution. We need to evaluate its molar conductivity ($\Lambda$).

Step 2: Key Formula or Approach:
The mathematical formula relating molar conductivity to specific conductivity when concentration is expressed in moles per liter (M) and conductivity is in $\text{S\ cm}^{-1}$ is:
$$\Lambda = \frac{1000 \times \kappa}{C}$$ Where $\kappa$ is the conductivity and $C$ is the molarity of the solution.

Step 3: Detailed Explanation:
Let's organize the values provided:
Conductivity ($\kappa$) = $2.67 \times 10^{-4}\ \text{S\ cm}^{-1}$
Molar Concentration ($C$) = $0.012\ \text{M}$
Substitute these metrics directly into our formulation:
$$\Lambda = \frac{1000 \times 2.67 \times 10^{-4}}{0.012}$$ Simplify the numerator:
$$1000 \times 2.67 \times 10^{-4} = 2.67 \times 10^{-1} = 0.267$$ Now substitute back to solve for $\Lambda$:
$$\Lambda = \frac{0.267}{0.012}$$ To eliminate the decimals, multiply the top and bottom expressions by 1000:
$$\Lambda = \frac{267}{12}$$ Performing standard long division:
$$\Lambda = 22.25\ \text{S\ cm}^2\ \text{mol}^{-1}$$ Rounding to one decimal digit as seen in the options yields $22.2\ \text{S\ cm}^2\ \text{mol}^{-1}$, which precisely matches option (D).

Step 4: Final Answer:
The molar conductivity of the solution is $22.2\ \text{S\ cm}^2\ \text{mol}^{-1}$, corresponding to option (D).