Question

What is molar conductivity of \(\text{CH}_3\text{COOH}\) at zero concentration if the molar conductivities of \(\text{H}_2\text{SO}_4\), \(\text{K}_2\text{SO}_4\) and \(\text{CH}_3\text{COOK}\) at zero concentrations are respectively x, y and z \(\Omega ^{-1}\text{cm}^2\text{mol}^{-1}\) ?

• A. \(\frac{(x-y)}{2}+z\)
• B. \((x-y+2z)\)
• C. \((x+y-z)\)
• D. \(\frac{(x-y)}{2}+2z\)

Hint:

Apply Kohlrausch law: write acetic acid as a combination of the three given electrolytes.

Answer 1

The Correct Option is A

Step 1: Goal
We want $\lambda_{\text{H}^+} + \lambda_{\text{Ac}^-}$ from the data.

Step 2: Per-ion parts
Half of the $\text{H}_2\text{SO}_4$ value minus half of the $\text{K}_2\text{SO}_4$ value removes the sulphate: $(x - y)/2 = \lambda_{\text{H}^+} - \lambda_{\text{K}^+}$.

Step 3: Add acetate
Adding the acetate value $z = \lambda_{\text{Ac}^-} + \lambda_{\text{K}^+}$ cancels the potassium term.

Step 4: Result
Sum is $\lambda_{\text{H}^+} + \lambda_{\text{Ac}^-} = (x-y)/2 + z$. The other options have wrong factors of 2 or wrong signs, and so give extra K or sulphate terms.

Final Answer:
The answer is (x-y)/2 + z. This is option (A). \[ \boxed{\text{(A) }\frac{x-y}{2}+z} \]