The freezing point depression constant for wa | vedclass

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(B) The formula for freezing point depression is $\Delta T_f = i 	imes K_f 	imes m$,where $m$ is the molality of the solution.First,calculate the mo

Vedclass · Accepted Answer

$2.63$

Vedclass · Answer

(B) The formula for freezing point depression is $\Delta T_f = i 	imes K_f 	imes m$,where $m$ is the molality of the solution.First,calculate the molar mass of $Na_2SO_4$: $M = (2 	imes 23) + 32 + (4 	imes 16) = 142 \, g \, mol^{-1}$.Calculate the molality $(m)$: $m = \frac{W_{solute} 	imes 1000}{M_{solute} 	imes W_{solvent}} = \frac{5.00 	imes 1000}{142 	imes 45.0} \approx 0.7824 \, m$.Using the formula $\Delta T_f = i 	imes K_f 	imes m$:$3.82 = i 	imes 1.86 	imes 0.7824$.Solving for $i$:$i = \frac{3.82}{1.86 	imes 0.7824} \approx \frac{3.82}{1.455} \approx 2.63$.

The freezing point depression constant for water is $1.86 \, ^\circ C \, m^{-1}$. If $5.00 \, g$ of $Na_2SO_4$ is dissolved in $45.0 \, g$ of $H_2O$,the freezing point is changed by $3.82 \, ^\circ C$. Calculate the van't Hoff factor $(i)$ for $Na_2SO_4$.

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