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The freezing point depression constant for water is $-1.86^\circ C\ m^{-1}$. If 5.00 g $Na_2SO_4$ is dissolved in 45.0 g $H_2O$, the freezing point is changed by $-3.82^\circ C$. Calculate the van't Hoff factor for $Na_2SO_4$.
A
0.381
B
2.05
C
2.63
D
3.11
Detailed Solution
$\Delta T_f = i\times K_f\times m$
Molar mass of $Na_2SO_4$ = 2(23) + 32 + 4(16) = 142 g $mol^{-1}$
Moles of $Na_2SO_4$ = $\frac{5.00}{142} = 0.0352$
Molality $m = \frac{0.0352}{45.0}\times1000 = 0.7825$ mol $kg^{-1}$
$i = \frac{\Delta T_f}{K_f\times m} = \frac{3.82}{1.86\times0.7825}$
$i = \frac{3.82}{1.455} = 2.63$
Molar mass of $Na_2SO_4$ = 2(23) + 32 + 4(16) = 142 g $mol^{-1}$
Moles of $Na_2SO_4$ = $\frac{5.00}{142} = 0.0352$
Molality $m = \frac{0.0352}{45.0}\times1000 = 0.7825$ mol $kg^{-1}$
$i = \frac{\Delta T_f}{K_f\times m} = \frac{3.82}{1.86\times0.7825}$
$i = \frac{3.82}{1.455} = 2.63$