Question

The freezing point depression constant for water is $-1.86^{\circ} Cm^{-1}$. 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

Answer 1

The Correct Option is C

To determine the Van't Hoff factor for Na_2SO_4, we will use the formula for freezing point depression:

\Delta T_f = i \cdot K_f \cdot m, where:

• \Delta T_f is the change in freezing point, given as -3.82^\circ C.
• K_f is the freezing point depression constant for water, 1.86^\circ C\, m^{-1}.
• m is the molality of the solution.
• i is the Van't Hoff factor.

First, calculate the molality (m) of the Na_2SO_4 solution. Molality is defined as moles of solute per kilogram of solvent:

Molecular weight of Na_2SO_4 = 2 \times 23 + 32 + 4 \times 16 = 142 \, g/mol.

Moles of Na_2SO_4 = \frac{5.00 \, \text{g}}{142 \, \text{g/mol}} = 0.0352 \, \text{mol}.

The mass of H_2O is 45.0 \, \text{g} = 0.0450 \, \text{kg}.

Molality (m) = \frac{0.0352 \, \text{mol}}{0.0450 \, \text{kg}} = 0.782 \, \text{mol/kg}.

Now substitute the values into the freezing point equation:

3.82 = i \times 1.86 \times 0.782

Solving for i,

i = \frac{3.82}{1.86 \times 0.782} = 2.63

Thus, the Van't Hoff factor for Na_2SO_4 is 2.63.