0.8 mL of acetic acid of density 1.06 g mL | vedclass

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(A) $1$. Calculate the mass of acetic acid $(CH_3COOH)$: $	ext{Mass} = 	ext{Density} 	imes 	ext{Volume} = 1.06 \ g \ mL^{-1} 	imes 0.8 \ mL = 0.8

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$1.24$

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(A) $1$. Calculate the mass of acetic acid $(CH_3COOH)$: $	ext{Mass} = 	ext{Density} 	imes 	ext{Volume} = 1.06 \ g \ mL^{-1} 	imes 0.8 \ mL = 0.848 \ g$.$2$. Calculate the moles of acetic acid: $	ext{Molar mass of } CH_3COOH = 60 \ g \ mol^{-1}$. $	ext{Moles} = \frac{0.848 \ g}{60 \ g \ mol^{-1}} \approx 0.01413 \ mol$.$3$. Calculate the molality $(m)$: Since the solvent is $1 \ kg$ of water,$m = 0.01413 \ mol \ kg^{-1}$.$4$. Calculate the theoretical depression in freezing point $(\Delta T_f)$: $\Delta T_f = K_f 	imes m = 1.86 \ K \ kg \ mol^{-1} 	imes 0.01413 \ mol \ kg^{-1} \approx 0.02628 \ K$.$5$. Calculate the Van't Hoff factor $(i)$: $i = \frac{\Delta T_f 	ext{ (observed)}}{\Delta T_f 	ext{ (theoretical)}} = \frac{0.0325}{0.02628} \approx 1.236 \approx 1.24$.

$0.8 \ mL$ of acetic acid of density $1.06 \ g \ mL^{-1}$ when dissolved in $1 \ kg$ of water causes a depression in freezing point by $0.0325^{\circ} C$. The Van't Hoff factor is:
$(K_{f} \text{ of } H_2O = 1.86 \ K \ kg \ mol^{-1})$

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$0.8 \ mL$ of acetic acid of density $1.06 \ g \ mL^{-1}$ when dissolved in $1 \ kg$ of water causes a depression in freezing point by $0.0325^{\circ} C$. The Van't Hoff factor is:$(K_{f} \text{ of } H_2O = 1.86 \ K \ kg \ mol^{-1})$