Question

Molar conductivity of $0.04\ \mathrm{M}$ $\mathrm{BaCl}_2$ solution is $230\ \Omega^{-1}\ \mathrm{cm}^2\ \mathrm{mol}^{-1}$ at $27^{\circ}\mathrm{C}$. What is its conductivity?

• A. $2.3 \times 10^{-3}\ \Omega^{-1}\ \mathrm{cm}^{-1}$
• B. $9.2 \times 10^{-3}\ \Omega^{-1}\ \mathrm{cm}^{-1}$
• C. $6.9 \times 10^{-3}\ \Omega^{-1}\ \mathrm{cm}^{-1}$
• D. $4.6 \times 10^{-3}\ \Omega^{-1}\ \mathrm{cm}^{-1}$

Hint:

To avoid messing up decimal places or getting confused by factors of 1000, convert your units carefully or remember the shortcut: $\kappa = \Lambda_m \times c \times 10^{-3}$. Multiplying $230 \times 0.04$ yields $9.2$, so matching it with the scaling factor gives $9.2 \times 10^{-3}$ instantly!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem asks for the electrolytic conductivity ($\kappa$) of a barium chloride ($\mathrm{BaCl}_2$) solution, given its molar concentration ($c = 0.04\ \mathrm{M}$) and its molar conductivity ($\Lambda_m = 230\ \Omega^{-1}\ \mathrm{cm}^2\ \mathrm{mol}^{-1}$).

Step 2: Key Formula or Approach:
The standard formula relating molar conductivity ($\Lambda_m$) to electrolytic conductivity ($\kappa$) when concentration ($c$) is expressed in $\mathrm{mol\ L}^{-1}$ ($\mathrm{M}$) and conductivity is in $\mathrm{S\ cm}^{-1}$ (or $\Omega^{-1}\ \mathrm{cm}^{-1}$) is: $$\Lambda_m = \frac{1000 \times \kappa}{c}$$ Rearranging this formula to explicitly isolate and solve for conductivity ($\kappa$) yields: $$\kappa = \frac{\Lambda_m \times c}{1000}$$

Step 3: Detailed Explanation:
Substitute the given numerical parameters directly into the rearranged equation: $$\Lambda_m = 230\ \Omega^{-1}\ \mathrm{cm}^2\ \mathrm{mol}^{-1}$$ $$c = 0.04\ \mathrm{mol\ L}^{-1}$$ $$\kappa = \frac{230 \times 0.04}{1000}$$ First, calculate the product in the numerator: $$230 \times 0.04 = 9.2$$ Now, perform the division by 1000 to complete the operation: $$\kappa = \frac{9.2}{1000} = 0.0092 = 9.2 \times 10^{-3}\ \Omega^{-1}\ \mathrm{cm}^{-1}$$

Step 4: Final Answer:
The conductivity of the solution is $9.2 \times 10^{-3}\ \Omega^{-1}\ \mathrm{cm}^{-1}$, which corresponds perfectly to option (B).