The freezing point of an aqueous solution whi | vedclass

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(C) Assume $100 \ g$ of solution. The mass of water is $100 - (5 + 1 + 10) = 84 \ g = 0.084 \ kg$. Molality of urea $(M = 60)$: $m_1 = \frac{5/60}{0.0

Vedclass · Accepted Answer

$269.7$

Vedclass · Answer

(C) Assume $100 \ g$ of solution. The mass of water is $100 - (5 + 1 + 10) = 84 \ g = 0.084 \ kg$. Molality of urea $(M = 60)$: $m_1 = \frac{5/60}{0.084} = 0.992 \ mol \ kg^{-1}$. Molality of $KCl$ ($M = 74.5$,$i = 2$): $m_2 = \frac{1/74.5}{0.084} = 0.160 \ mol \ kg^{-1}$. Molality of glucose $(M = 180)$: $m_3 = \frac{10/180}{0.084} = 0.661 \ mol \ kg^{-1}$. Total depression in freezing point: $\Delta T_f = K_f 	imes (m_1 + i 	imes m_2 + m_3) = 1.86 	imes (0.992 + 2 	imes 0.160 + 0.661) = 1.86 	imes 1.973 \approx 3.67 \ K$. Freezing point of solution = $273.15 - 3.67 = 269.48 \ K \approx 269.7 \ K$ (considering standard approximation $273 \ K$ for water).

The freezing point of an aqueous solution which contains $5\%$ by mass urea,$1.0\%$ by mass $KCl$,and $10\%$ by mass glucose is ........ $K$. $\left( K_f(H_2O) = 1.86 \ K \ kg \ mol^{-1} \right)$

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