Question

The conductivity of $0.04\ \mathrm{M}\ \mathrm{BaCl_2}$ solution is $0.0112\ \Omega^{-1}\ \mathrm{cm^{-1}}$ at $25^\circ\text{C}$. What is its molar conductivity?

• A. $357.0\ \Omega^{-1}\ \mathrm{cm^2\ mol^{-1}}$
• B. $140.0\ \Omega^{-1}\ \mathrm{cm^2\ mol^{-1}}$
• C. $44.8\ \Omega^{-1}\ \mathrm{cm^2\ mol^{-1}}$
• D. $280.0\ \Omega^{-1}\ \mathrm{cm^2\ mol^{-1}}$

Hint:

When using the equation $\Lambda_m = \frac{\kappa \times 1000}{c}$, make sure $\kappa$ is in $\mathrm{cm^{-1}}$ units. To make the division easier, clear out decimals by expanding the fraction: $\frac{11.2}{0.04} = \frac{1120}{4} = 280$. This keeps your mental calculations fast and error-free!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given the molar concentration ($c = 0.04\ \mathrm{M}$) and the specific electrolytic conductivity ($\kappa = 0.0112\ \mathrm{\Omega^{-1}\ cm^{-1}}$) of a barium chloride ($\mathrm{BaCl_2}$) solution. We need to calculate its molar conductivity ($\Lambda_m$) in standard units of $\mathrm{\Omega^{-1}\ cm^2\ mol^{-1}}$.

Step 2: Key Formula or Approach:
The standard equation for calculating molar conductivity when conductivity ($\kappa$) is expressed in $\mathrm{\Omega^{-1}\ cm^{-1}}$ and concentration ($c$) is in $\mathrm{mol\ L^{-1}}$ ($\mathrm{M}$) is given by: $$\Lambda_m = \frac{\kappa \times 1000}{c}$$

Step 3: Detailed Explanation:
Substitute our given values into the molar conductivity formula: $$\Lambda_m = \frac{0.0112\ \Omega^{-1}\ cm^{-1} \times 1000\ cm^3\ L^{-1}}{0.04\ mol\ L^{-1}}$$ Simplify the numerator value: $$0.0112 \times 1000 = 11.2$$ Now, perform the division: $$\Lambda_m = \frac{11.2}{0.04} = \frac{1120}{4} = 280\ \Omega^{-1}\ cm^2\ mol^{-1}$$

Step 4: Final Answer:
The molar conductivity of the solution is $280.0\ \Omega^{-1}\ \mathrm{cm^2\ mol^{-1}}$, which corresponds to option (D).