Question

The molar conductivity of \(0.04\) M \(\text{AB}_2\) type salt solution at \(300\) K is \(200\,\Omega ^{-1}\text{cm}^2\text{mol}^{-1}\). Find the conductivity.

• A. \(0.006\,\Omega ^{-1}\text{cm}^{-1}\)
• B. \(0.008\,\Omega ^{-1}\text{cm}^{-1}\)
• C. \(0.01\,\Omega ^{-1}\text{cm}^{-1}\)
• D. \(0.015\,\Omega ^{-1}\text{cm}^{-1}\)

Hint:

kappa = Lambda_m x C / 1000.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Conductivity \(\kappa\) and molar conductivity \(\Lambda_m\) are linked by concentration. With \(C\) in mol/L and \(\kappa\) in \(\Omega^{-1}cm^{-1}\),
\[ \Lambda_m = \frac{1000\,\kappa}{C} \]

Step 2: Rearrange:
\[ \kappa = \frac{\Lambda_m C}{1000} \]

Step 3: Substitute:
\[ \kappa = \frac{200\times 0.04}{1000} = \frac{8}{1000} = 0.008\ \Omega^{-1}\text{cm}^{-1} \]

Step 4: Note:
The salt type \(AB_2\) and the temperature are not needed, because \(\Lambda_m\) is already given for the salt as a whole. The answer is option (B).

Final Answer:
The conductivity is 0.008 per ohm per cm. \[ \boxed{0.008\ \Omega^{-1}\text{cm}^{-1}} \]