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: Note the given quantities.
Concentration $c = 0.04\ M$ and conductivity $\kappa = 0.0112\ \Omega^{-1}\,cm^{-1}$.
Step 2: Recall the molar conductivity formula.
When $\kappa$ is in $\Omega^{-1}\,cm^{-1}$ and $c$ is in $mol\,L^{-1}$, \[ \Lambda_m = \frac{1000\,\kappa}{c}. \] The factor $1000$ converts litres to cubic centimetres.
Step 3: Build the numerator.
$1000 \times 0.0112 = 11.2$.
Step 4: Divide by the concentration.
\[ \Lambda_m = \frac{11.2}{0.04} = \frac{1120}{4} = 280. \]
Step 5: Attach the units.
The result carries units $\Omega^{-1}\,cm^{2}\,mol^{-1}$, so $\Lambda_m = 280\ \Omega^{-1}\,cm^{2}\,mol^{-1}$.
Step 6: Conclude.
The molar conductivity is $280.0\ \Omega^{-1}\,cm^{2}\,mol^{-1}$, option (D).
\[ \boxed{280.0\ \Omega^{-1}\,cm^{2}\,mol^{-1} \text{ (option D)}} \]