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:
Rupal, Shanu and Trisha were partners in a firm sharing profits and losses in the ratio of 4:3:1. Their Balance Sheet as at 31st March, 2024 was as follows:
(i) Trisha's share of profit was entirely taken by Shanu.
(ii) Fixed assets were found to be undervalued by Rs 2,40,000.
(iii) Stock was revalued at Rs 2,00,000.
(iv) Goodwill of the firm was valued at Rs 8,00,000 on Trisha's retirement.
(v) The total capital of the new firm was fixed at Rs 16,00,000 which was adjusted according to the new profit sharing ratio of the partners. For this necessary cash was paid off or brought in by the partners as the case may be.
Prepare Revaluation Account and Partners' Capital Accounts.