To find the molal freezing-point constant ([tex]\(K_f\)[/tex]) of the unknown solvent, we can use the formula related to freezing point depression, which is given by:
[tex]\[ \Delta T_f = K_f \cdot m \][/tex]
where:
- [tex]\(\Delta T_f\)[/tex] is the freezing-point depression. For this problem, [tex]\(\Delta T_f\)[/tex] is 7.8°C.
- [tex]\(K_f\)[/tex] is the molal freezing-point constant of the solvent.
- [tex]\(m\)[/tex] is the molality of the solution.
First, let's determine the molality ([tex]\(m\)[/tex]) of the solution.
Molality ([tex]\(m\)[/tex]) is defined as the number of moles of solute per kilogram of solvent:
[tex]\[ m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \][/tex]
In the given problem:
- The number of moles of solute is 2 moles.
- The mass of the solvent is 1 kg.
Therefore,
[tex]\[ m = \frac{2 \, \text{moles}}{1 \, \text{kg}} = 2 \, \text{mol/kg} \][/tex]
Next, we need to solve for the molal freezing-point constant ([tex]\(K_f\)[/tex]). Rearranging the freezing-point depression formula to solve for [tex]\(K_f\)[/tex], we get:
[tex]\[ K_f = \frac{\Delta T_f}{m} \][/tex]
Substituting the known values:
[tex]\[ K_f = \frac{7.8^\circ \text{C}}{2 \, \text{mol/kg}} = 3.9^\circ \text{C/mol/kg} \][/tex]
Therefore, the molal freezing-point constant of the unknown solvent is [tex]\(3.9^\circ \text{C/mol/kg}\)[/tex].
The correct answer is:
c. [tex]\(-3.9^\circ \text{C/mol/kg}\)[/tex]
