6. (2018 T1) Two moles of a nonelectrolyte solute are dissolved in 1 kg of an unknown solvent. The solution freezes at [tex]$7.8^{\circ} C$[/tex] below its normal freezing point. What is the molal freezing-point constant of the unknown solvent?

A. [tex]$-14.5^{\circ} C / m$[/tex]
B. [tex]$-5.08^{\circ} C / m$[/tex]
C. [tex]$-3.9^{\circ} C / m$[/tex]
D. [tex]$-1.86^{\circ} C / m$[/tex]



Answer :

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]