An aqueous solution contains 6% and 4% by wei | vedclass

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(A) Consider $100 \, g$ of the solution. The mass of $HA = 6 \, g$ and the mass of urea = $4 \, g$.The mass of water = $100 - (6 + 4) = 90 \, g$.For $

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$0.9$

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(A) Consider $100 \, g$ of the solution. The mass of $HA = 6 \, g$ and the mass of urea = $4 \, g$.The mass of water = $100 - (6 + 4) = 90 \, g$.For $HA$,$n_1 = \frac{6}{60} = 0.1 \, mol$. For urea,$n_2 = \frac{4}{60} = 0.0667 \, mol$.Using the formula $\Delta T_f = K_f 	imes 	ext{molality} 	imes i_{total}$,where $i_{total} = \frac{i_1 n_1 + i_2 n_2}{n_1 + n_2}$.$\Delta T_f = K_f 	imes \frac{(i_1 n_1 + i_2 n_2) 	imes 1000}{W_{solvent(g)}}$.$\frac{13.6}{3} = 1.8 	imes \frac{((1+\alpha) 	imes 0.1 + 1 	imes 0.0667) 	imes 1000}{90}$.$\frac{13.6}{3} = 20 	imes (0.1 + 0.1\alpha + 0.0667) = 20 	imes (0.1667 + 0.1\alpha)$.$4.533 = 3.334 + 2\alpha$ $\Rightarrow 2\alpha = 1.199$ $\Rightarrow \alpha \approx 0.6$.$K_a = \frac{C\alpha^2}{1-\alpha}$. Here $C = \frac{0.1 \, mol}{0.09 \, L} = 1.11 \, M$.$K_a = \frac{1.11 	imes (0.6)^2}{1-0.6} = \frac{1.11 	imes 0.36}{0.4} \approx 1$.

An aqueous solution contains $6\%$ and $4\%$ by weight of a weak acid $(HA)$ and urea respectively. If the depression in freezing point is $\frac{13.6}{3} \, ^{\circ}C$,calculate the $K_a$ of $HA$. [Given: $K_f = 1.8 \, K \, kg/mol$,Molecular weight of $HA = 60$,$d_{solution} = 1 \, g/mL$]

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