Calculate the van't Hoff factor (i) for an aq | vedclass

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Question

(A) Step $1$: Calculate the depression in freezing point $(\Delta T_f)$.$\Delta T_f = T_f^{\circ} - T_f = 0^{\circ}	ext{C} - (-0.045^{\circ}	ext{C})

Vedclass · Accepted Answer

$1.21$

Vedclass · Answer

(A) Step $1$: Calculate the depression in freezing point $(\Delta T_f)$.$\Delta T_f = T_f^{\circ} - T_f = 0^{\circ}	ext{C} - (-0.045^{\circ}	ext{C}) = 0.045 	ext{ K}$.Step $2$: Use the formula for depression in freezing point: $\Delta T_f = i 	imes K_f 	imes m$.Step $3$: Substitute the given values: $0.045 = i 	imes 1.86 	imes 0.02$.Step $4$: Solve for $i$: $i = \frac{0.045}{1.86 	imes 0.02} = \frac{0.045}{0.0372} \approx 1.21$.

Calculate the van't Hoff factor $(i)$ for an aqueous solution of $0.02 \text{ m}$ formic acid if it freezes at $-0.045^{\circ}\text{C}$. Given: $K_f = 1.86 \text{ K kg mol}^{-1}$ and the freezing point of pure water $= 0^{\circ}\text{C}$.

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