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: Connect the two quantities.
Molar conductivity and specific conductivity are linked by \[ \Lambda = \frac{1000\,k}{C} \]

Step 2: Solve for k.
Rearrange to get $k = \dfrac{\Lambda\,C}{1000}$.

Step 3: Put in the numbers.
With $\Lambda = 106$ and $C = 0.1$, \[ k = \frac{106 \times 0.1}{1000} = 1.06 \times 10^{-2}\ \Omega^{-1}\,\text{cm}^{-1} \]
\[ \boxed{1.06 \times 10^{-2}\ \Omega^{-1}\,\text{cm}^{-1},\ \text{option 1}} \]