Question

The molar conductivity of 0.1 M $\text{BaCl}_2$ solution is $106\ \Omega^{-1}\ \text{cm}^2\ \text{mol}^{-1}$ at $25^\circ\text{C}$. What is its conductivity?

• A. $1.06 \times 10^{-2}\ \Omega^{-1}\ \text{cm}^{-1}$
• B. $5.3 \times 10^{-3}\ \Omega^{-1}\ \text{cm}^{-1}$
• C. $3.66 \times 10^{-3}\ \Omega^{-1}\ \text{cm}^{-1}$
• D. $2.6 \times 10^{-2}\ \Omega^{-1}\ \text{cm}^{-1}$

Hint:

Be mindful of units! Multiplying molarity by 0.1 is equivalent to shifting the decimal place left by one position. Dividing by 1000 shifts it three positions further, matching $1.06 \times 10^{-2}$ perfectly.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given the molar conductivity ($\wedge$) and the molar concentration ($C$) of a barium chloride ($\text{BaCl}_2$) solution. We need to determine its specific conductivity ($k$, kappa).

Step 2: Key Formula or Approach:
The standard formula linking molar conductivity to specific conductivity when concentration is given in $\text{mol L}^{-1}$ and conductivity is in $\Omega^{-1}\ \text{cm}^{-1}$ is: $$ \wedge = \frac{1000 \times k}{C} $$ Rearranging this equation to solve for specific conductivity ($k$) gives: $$ k = \frac{\wedge \times C}{1000} $$

Step 3: Detailed Explanation:
Let's substitute the given parameters into our rearranged expression:

• Molar conductivity, $\wedge = 106\ \Omega^{-1}\ \text{cm}^2\ \text{mol}^{-1}$

• Molarity, $C = 0.1\ \text{M} = 0.1\ \text{mol L}^{-1}$
Calculating the value of $k$: $$ k = \frac{106 \times 0.1}{1000} $$ $$ k = \frac{10.6}{1000} $$ $$ k = 1.06 \times 10^{-2}\ \Omega^{-1}\ \text{cm}^{-1} $$

Step 4: Final Answer:
The conductivity of the solution is $1.06 \times 10^{-2}\ \Omega^{-1}\ \text{cm}^{-1}$, matching option (A).