Question

What is the molar conductivity of $0.05\ \text{M}$ solution of sodium hydroxide, if its conductivity is $0.0118\ \text{S}\ \text{cm}^{-1}$ at $298\ \text{K}$?

• A. $236\ \text{S}\ \text{cm}^2\ \text{mol}^{-1}$
• B. $423\ \text{S}\ \text{cm}^2\ \text{mol}^{-1}$
• C. $354\ \text{S}\ \text{cm}^2\ \text{mol}^{-1}$
• D. $590\ \text{S}\ \text{cm}^2\ \text{mol}^{-1}$

Hint:

Dividing a value by $0.05$ is mathematically identical to multiplying that value directly by 20. Shifting your numerator to $11.8$ and multiplying by 20 ($11.8 \times 2 = 23.6$, then scaled by 10) gives $236$ instantly.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem requires computing the molar conductivity ($\wedge_m$) of a sodium hydroxide (NaOH) solution given its molar concentration ($c$) and its experimental electrolytic conductivity ($\kappa$).

Step 2: Key Formula or Approach:
The mathematical relationship linking molar conductivity to specific conductivity when units are expressed in terms of centimeters and liters is given by:
$$ \wedge_m = \frac{1000 \times \kappa}{c} $$ Here, $\kappa$ represents the specific conductivity in $\text{S}\ \text{cm}^{-1}$ and $c$ is the molarity in $\text{mol}\ \text{L}^{-1}$.

Step 3: Detailed Explanation:
Let's substitute the given numerical constants directly into our operational equation:
Conductivity ($\kappa$) $= 0.0118\ \text{S}\ \text{cm}^{-1}$
Molar concentration ($c$) $= 0.05\ \text{M} = 0.05\ \text{mol}\ \text{L}^{-1}$
$$ \wedge_m = \frac{1000 \times 0.0118}{0.05} $$ Evaluating the numerator product shifts the decimal three places to the right:
$$ \wedge_m = \frac{11.8}{0.05} $$ To simplify the division step, multiply both the numerator and denominator parameters by 100:
$$ \wedge_m = \frac{1180}{5} = 236\ \text{S}\ \text{cm}^2\ \text{mol}^{-1} $$

Step 4: Final Answer:
The evaluated molar conductivity value is $236\ \text{S}\ \text{cm}^2\ \text{mol}^{-1}$, which corresponds directly to option (A).