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: Understanding the Question:
The question asks for the molar conductivity ($\Lambda_m$) of a potassium chloride (KCl) solution, given its molar concentration ($M$) and its specific electrolytic conductivity ($\kappa$).

Step 2: Key Formula or Approach:
The mathematical relationship linking molar conductivity to electrolytic conductivity and molarity is: $$\Lambda_m = \frac{\kappa \times 1000}{M}$$ where: $\kappa$ is the conductivity in $\text{S cm}^{-1}$ $M$ is the molarity of the solution in $\text{mol L}^{-1}$ The factor of 1000 converts liters (L or $\text{dm}^3$) into cubic centimeters ($\text{cm}^3$).

Step 3: Detailed Explanation:
Given parameters: $\kappa = 0.0627\text{ S cm}^{-1}$ $M = 0.3\text{ M} = 0.3\text{ mol L}^{-1}$ Substitute the variables into the formula: $$\Lambda_m = \frac{0.0627 \times 1000}{0.3}$$ $$\Lambda_m = \frac{62.7}{0.3}$$ $$\Lambda_m = 209\text{ S cm}^2\text{ mol}^{-1}$$

Step 4: Final Answer:
The molar conductivity of the solution is 209 S cm$^2$ mol$^{-1}$, which matches option (C).