If the Van’t Hoff Factor for NaCl is 1.88, what is the freezing point of a 0.50 molal NaCl solution in water? Kf= 1.86 degrees C/m for water.
a. -0.93 degrees C b. -1.75 degrees C c. -1.86 degrees C d. 1.86 degrees C
Tf= 0 – (i)(Kf)(m)
Tf= 0 – (1.88) (1.86 degrees C/m) (0.5 m) Tf= – 1.75 degrees C
The correct answer and explanation is :
The freezing point depression formula is:
[
\Delta T_f = i \cdot K_f \cdot m
]
where:
- (\Delta T_f) is the change in freezing point.
- (i) is the Van’t Hoff factor, which accounts for the number of particles produced when a solute dissolves. For NaCl, (i = 1.88), because NaCl dissociates into Na(^+) and Cl(^-) ions in solution.
- (K_f) is the cryoscopic constant, also known as the freezing point depression constant. For water, (K_f = 1.86 \, \text{°C/m}).
- (m) is the molality of the solution, which is given as 0.50 molal.
Now, applying the formula:
[
\Delta T_f = (1.88) \cdot (1.86 \, \text{°C/m}) \cdot (0.50 \, \text{mol/kg})
]
[
\Delta T_f = 1.75 \, \text{°C}
]
The freezing point depression (\Delta T_f) is 1.75°C. Since freezing point depression is the amount by which the freezing point decreases, we subtract this value from the freezing point of pure water (which is 0°C):
[
T_f = 0 \, \text{°C} – 1.75 \, \text{°C} = -1.75 \, \text{°C}
]
Thus, the correct answer is b. -1.75°C.
Explanation:
The Van’t Hoff factor (i) tells us how many particles are produced when the solute dissolves. For NaCl, it dissociates into two ions, Na(^+) and Cl(^-), and the factor accounts for this dissociation. The molality ((m)) is the number of moles of solute per kilogram of solvent, which is 0.50 molal in this case. The cryoscopic constant (K_f) is a property of the solvent (water) that tells us how much the freezing point will drop per molal concentration of solute. Combining these values gives the change in freezing point, and from that, we subtract from the original freezing point of water to get the final freezing point of the solution.