The freezing point depression constant for wa | vedclass

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(B) The formula for freezing point depression is $\Delta T_f = i 	imes K_f 	imes m.$ Here,$\Delta T_f = 3.82 \, ^oC,$ $K_f = 1.86 \, ^oC \, kg \, mo

Vedclass · Accepted Answer

$2.63$

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(B) The formula for freezing point depression is $\Delta T_f = i 	imes K_f 	imes m.$ Here,$\Delta T_f = 3.82 \, ^oC,$ $K_f = 1.86 \, ^oC \, kg \, mol^{-1},$ $W_{solute} = 5.00 \, g,$ $W_{solvent} = 45.0 \, g,$ and the molar mass of $Na_2SO_4$ is $M = 142 \, g \, mol^{-1}.$ The molality $m$ is given by $\frac{W_{solute} 	imes 1000}{M 	imes W_{solvent}} = \frac{5.00 	imes 1000}{142 	imes 45.0} \approx 0.782 \, m.$ Substituting these values into the equation: $3.82 = i 	imes 1.86 	imes 0.782.$ Solving for $i$: $i = \frac{3.82}{1.86 	imes 0.782} \approx 2.63.$

The freezing point depression constant for water is $1.86 \, ^oC \, kg \, mol^{-1}.$ If $5.00 \, g$ of $Na_2SO_4$ is dissolved in $45.0 \, g$ of $H_2O,$ the freezing point is lowered by $3.82 \, ^oC.$ Calculate the van't Hoff factor for $Na_2SO_4.$

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