Electrochemistry

Hint: During electrolysis, of dilute H₂SO₄, at anode, 2H₂O(l) → O₂ (g) + 4H⁺ (aq) + 4e^–

Explanation:
The reason (R) provides the molecular weight of oxygen. The calculation for the volume of oxygen evolved in (A) depends on Faraday’s law and the volume occupied by one mole of gas at STP (22.4 L/mol), which is a consequence of Avogadro’s Law and the Ideal Gas Law. While the molecular weight (32) is consistent with \(O_2​\) occupying 22.4 L/mol at STP, it is not the fundamental chemical or physical principle required to calculate the 11.2 L value based on the 2 F charge. The calculation primarily relies on the stoichiometry of the half-reaction and Faraday’s Laws.

Both (A) and (R) are True, but (R) is not the correct explanation of (A)

 

Hint: $\frac{{\mathrm{\Lambda }}_{{\mathrm{CH}}_3\mathrm{COOH}}}{\mathrm{\Lambda }{°}_{{\mathrm{CH}}_3\mathrm{COOH}}}=\mathrm{\alpha }$ 

Calculate the value  $\mathrm{\Lambda }{°}_{{\mathrm{CH}}_3\mathrm{COOH}}$ as follows:

$\mathrm{\Lambda }{°}_{{\mathrm{CH}}_3\mathrm{COOH}}=\mathrm{\Lambda }{°}_{{\mathrm{H}}^{+}}+\mathrm{\Lambda }{°}_{{\mathrm{CH}}_3{\mathrm{COO}}^{–}}$
$\mathrm{\Lambda }{°}_{{\mathrm{CH}}_3\mathrm{COOH}}$ $=350$ $+$ $50$
$\mathrm{\Lambda }{°}_{{\mathrm{CH}}_3\mathrm{COOH}}$ $=$ $400$ $\mathrm{S}$ ${\mathrm{cm}}^2$ ${\mathrm{mol}}^{–1}$

Calculate the value of $\alpha$ as follows:

$\frac{{\mathrm{\Lambda }}_{{\mathrm{CH}}_3\mathrm{COOH}}}{\mathrm{\Lambda }{°}_{{\mathrm{CH}}_3\mathrm{COOH}}}=\mathrm{\alpha }$

$\frac{20}{400}=\mathrm{\alpha }$
$\mathrm{\alpha }=0.05$

Step 2:

Calculate the value of the dissociation constant as follows:

${\mathrm{CH}}_3\mathrm{COOH}\leftrightarrow {\mathrm{CH}}_3{\mathrm{COO}}^{–}$ $+$ ${\mathrm{H}}^{+}$
$\mathrm{C}$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $–$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $–$
$\mathrm{C}–\mathrm{C\alpha }$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $\mathrm{C\alpha }$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $\mathrm{C\alpha }$

${\mathrm{K}}_{\mathrm{a}}=\frac{\mathrm{C\alpha }\times \mathrm{C\alpha }}{\mathrm{C}(1–\mathrm{\alpha })}$
${\mathrm{K}}_{\mathrm{a}}=\frac{{\mathrm{C\alpha }}^2}{(1–\mathrm{\alpha })}$
${\mathrm{K}}_{\mathrm{a}}=\frac{0.007\times 0.05\times 0.05}{1–0.05}$
K_a   =  1.75×10^-5

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