A sample of water is found to contain 5.85 % | vedclass

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(A) $1$. Calculate molality $(m)$ of each solute in $100 \ g$ of solution (assuming $84.65 \ g$ of water solvent): $m_{AB} = \frac{5.85 \ g / 58.5 \ g

Vedclass · Accepted Answer

$264.25$

Vedclass · Answer

(A) $1$. Calculate molality $(m)$ of each solute in $100 \ g$ of solution (assuming $84.65 \ g$ of water solvent): $m_{AB} = \frac{5.85 \ g / 58.5 \ g \ mol^{-1}}{0.08465 \ kg} = 1.181 \ mol \ kg^{-1}$ $m_{XY_2} = \frac{9.50 \ g / 95 \ g \ mol^{-1}}{0.08465 \ kg} = 1.181 \ mol \ kg^{-1}$ $2$. Calculate van't Hoff factor $(i)$ for each: For $AB ightarrow A^+ + B^-$, $i_1 = 1 + (2-1)0.8 = 1.8$ For $XY_2 ightarrow X^{2+} + 2Y^-$, $i_2 = 1 + (3-1)0.6 = 2.2$ $3$. Calculate total depression in freezing point $(\Delta T_f)$: $\Delta T_f = K_f 	imes (i_1 m_1 + i_2 m_2) = 1.86 	imes (1.8 	imes 1.181 + 2.2 	imes 1.181) = 1.86 	imes (4.724) \approx 8.786 \ K$ $4$. Calculate final freezing point: $T_f = 273 \ K - 8.786 \ K = 264.214 \ K \approx 264.25 \ K$.

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