83 g of ethylene glycol is dissolved in 625 | vedclass

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(B) $1$. Calculate the molar mass of ethylene glycol $(C_2H_6O_2)$: $(2 	imes 12) + (6 	imes 1) + (2 	imes 16) = 62 \ g \ mol^{-1}$.$2$. Calculate

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$269$

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(B) $1$. Calculate the molar mass of ethylene glycol $(C_2H_6O_2)$: $(2 	imes 12) + (6 	imes 1) + (2 	imes 16) = 62 \ g \ mol^{-1}$.$2$. Calculate the number of moles of solute: $n = \frac{83 \ g}{62 \ g \ mol^{-1}} \approx 1.3387 \ mol$.$3$. Calculate the molality $(m)$ of the solution: $m = \frac{n 	ext{ (mol)}}{W_{	ext{solvent}} 	ext{ (kg)}} = \frac{1.3387 \ mol}{0.625 \ kg} = 2.1419 \ mol \ kg^{-1}$.$4$. Calculate the depression in freezing point $(\Delta T_f)$: $\Delta T_f = K_f 	imes m = 1.86 	imes 2.1419 \approx 3.984 \ K \approx 4 \ K$.$5$. Calculate the freezing point of the solution $(T_f)$: $T_f = T_f^{\circ} - \Delta T_f = 273 \ K - 4 \ K = 269 \ K$.

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