Calculate \( \Lambda_m^0 \) for acetic acid and its degree of dissociation (\( \alpha \)) if its molar conductivity is 48.1 \( \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \).
Given that
\( \Lambda_m^0 (\text{HC}) = 426 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \),
\( \Lambda_m^0 (\text{NaCl}) = 126 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \),
\( \Lambda_m^0 (\text{CH}_3\text{COONa}) = 91 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \).
To solve the problem, we need to calculate the limiting molar conductivity (\( \Lambda_m^0 \)) of acetic acid and its degree of dissociation (\( \alpha \)), given its molar conductivity (\( \Lambda_m \)) as 48.1 \( \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \), and the limiting molar conductivities: \( \Lambda_m^0 (\text{HCl}) = 426 \), \( \Lambda_m^0 (\text{NaCl}) = 126 \), and \( \Lambda_m^0 (\text{CH}_3\text{COONa}) = 91 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \).
1. Calculate \( \Lambda_m^0 \) for Acetic Acid:
Acetic acid (\( \text{CH}_3\text{COOH} \)) dissociates into \( \text{H}^+ \) and \( \text{CH}_3\text{COO}^- \). We use Kohlrausch’s law to find \( \Lambda_m^0 (\text{CH}_3\text{COOH}) \): \( \Lambda_m^0 (\text{CH}_3\text{COOH}) = \lambda_m^0 (\text{H}^+) + \lambda_m^0 (\text{CH}_3\text{COO}^-) \).
We can find these ionic conductivities using the given data:
- \( \Lambda_m^0 (\text{HCl}) = \lambda_m^0 (\text{H}^+) + \lambda_m^0 (\text{Cl}^-) = 426 \),
- \( \Lambda_m^0 (\text{NaCl}) = \lambda_m^0 (\text{Na}^+) + \lambda_m^0 (\text{Cl}^-) = 126 \),
- \( \Lambda_m^0 (\text{CH}_3\text{COONa}) = \lambda_m^0 (\text{Na}^+) + \lambda_m^0 (\text{CH}_3\text{COO}^-) = 91 \).
Subtract the second equation from the first to eliminate \( \lambda_m^0 (\text{Cl}^-) \):
\( \lambda_m^0 (\text{H}^+) + \lambda_m^0 (\text{Cl}^-) - (\lambda_m^0 (\text{Na}^+) + \lambda_m^0 (\text{Cl}^-)) = 426 - 126 \),
\( \lambda_m^0 (\text{H}^+) - \lambda_m^0 (\text{Na}^+) = 300 \).
Now use the third equation with \( \lambda_m^0 (\text{Na}^+) \):
From the third equation, \( \lambda_m^0 (\text{CH}_3\text{COO}^-) = 91 - \lambda_m^0 (\text{Na}^+) \).
We need \( \lambda_m^0 (\text{H}^+) + \lambda_m^0 (\text{CH}_3\text{COO}^-) \):
Add \( \lambda_m^0 (\text{H}^+) - \lambda_m^0 (\text{Na}^+) = 300 \) to \( \lambda_m^0 (\text{Na}^+) + \lambda_m^0 (\text{CH}_3\text{COO}^-) = 91 \):
\( \lambda_m^0 (\text{H}^+) + \lambda_m^0 (\text{CH}_3\text{COO}^-) = 300 + 91 = 391 \).
Thus, \( \Lambda_m^0 (\text{CH}_3\text{COOH}) = 391 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \).
2. Calculate the Degree of Dissociation (\( \alpha \)):
The degree of dissociation is given by \( \alpha = \frac{\Lambda_m}{\Lambda_m^0} \):
\( \alpha = \frac{48.1}{391} \approx 0.123 \).
Final Answer:
The limiting molar conductivity \( \Lambda_m^0 \) for acetic acid is \( 391 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \), and the degree of dissociation \( \alpha \) is approximately \( 0.123 \).
Europium (Eu) resembles Calcium (Ca) in the following ways:
(A). Both are diamagnetic
(B). Insolubility of their sulphates and carbonates in water
(C). Solubility of these metals in liquid NH3
(D). Insolubility of their dichlorides in strong HCI
Choose the correct answer from the options given below: