1.2 mL of acetic acid having density 1.06 g | vedclass

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Question

(B) The formula for depression in freezing point is $\Delta T_f = i 	imes K_f 	imes m$. Mass of acetic acid = $density 	imes volume = 1.06 \ g \ cm

Vedclass · Accepted Answer

$1.04$

Vedclass · Answer

(B) The formula for depression in freezing point is $\Delta T_f = i 	imes K_f 	imes m$. Mass of acetic acid = $density 	imes volume = 1.06 \ g \ cm^{-3} 	imes 1.2 \ mL = 1.272 \ g$. Moles of acetic acid = $\frac{1.272 \ g}{60 \ g \ mol^{-1}} = 0.0212 \ mol$. Since $1 \ L$ of water is used,the mass of the solvent is $1000 \ g = 1 \ kg$. Molality $(m) = \frac{0.0212 \ mol}{1 \ kg} = 0.0212 \ mol \ kg^{-1}$. Substituting the values into the formula: $0.041 = i 	imes 1.86 	imes 0.0212$. $i = \frac{0.041}{1.86 	imes 0.0212} \approx \frac{0.041}{0.039432} \approx 1.04$. Thus,the van't Hoff factor is $1.04$. Therefore,option $(B)$ is correct.

$1.2 \ mL$ of acetic acid having density $1.06 \ g \ cm^{-3}$ is dissolved in $1 \ L$ of water. The depression in freezing point observed for this concentration of acid was $0.041^{\circ} C$. The van't Hoff factor of the acid is $(K_f \text{ of water } = 1.86 \ K \ kg \ mol^{-1})$

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