Using Equation #1 on page 8 of the lab manual, what would be the change in freezing point for ethanol (Kf 1.99°C/m) if the molality of a dissolved solute was 0.420 m? (Assume a van Hoff value of
a) 0.420°C
b) 4.74°C
c) 0.8358°C
d) 1.99°C
The Correct Answer and Explanation is :
To calculate the change in freezing point for ethanol, we can use the equation for freezing point depression:
[
\Delta T_f = K_f \times m \times i
]
Where:
- (\Delta T_f) is the change in freezing point (in °C),
- (K_f) is the freezing point depression constant (in °C/m),
- (m) is the molality of the solution (in mol/kg), and
- (i) is the van’t Hoff factor, which accounts for the number of particles into which the solute dissociates.
Given:
- (K_f = 1.99 \, \text{°C/m}),
- (m = 0.420 \, \text{m}),
- (i = 1), since the solute is assumed to be non-electrolyte or does not dissociate into multiple ions.
Now, we can calculate the change in freezing point:
[
\Delta T_f = 1.99 \times 0.420 \times 1 = 0.8358 \, \text{°C}
]
Thus, the change in freezing point is (0.8358) °C.
Therefore, the correct answer is c) 0.8358°C.
Explanation:
Freezing point depression occurs when a solute is dissolved in a solvent, causing the freezing point of the solvent to decrease. The equation used here is derived from colligative properties, which depend on the number of solute particles in the solution, not on their chemical nature. The freezing point depression constant ((K_f)) is specific to each solvent. In this case, (K_f) for ethanol is given as 1.99°C/m, and the molality of the solution is 0.420 m. The van’t Hoff factor ((i)) is used to account for dissociation of the solute into multiple ions, but since the solute here does not dissociate, (i = 1). By multiplying these values together, we obtain the change in freezing point, which represents how much lower the freezing point of ethanol would be compared to pure ethanol.