Calculate the Freezing Point Depression (DT) for the following solution. (Assume the density of water is 1.00 g/mL and Kf = 1.86 Deg C/m)
3.4 M NaCl (Assume the density of the solution is 1.00 g/mL.)
- Calculate the Freezing Point Depression (DT) for the following solution. (Assume the density of water is 1.00 g/mL and Kf = 1.86 Deg C/m)
27g of KCl in 1.6 L of water
The Correct Answer and Explanation is:
To calculate the freezing point depression (ΔT_f) for a solution, we use the formula:
[
\Delta T_f = i \cdot K_f \cdot m
]
Where:
- i is the van’t Hoff factor, which represents the number of particles into which a solute dissociates in solution. For NaCl, (i = 2) (since NaCl dissociates into Na⁺ and Cl⁻ ions), and for KCl, (i = 2) (since KCl dissociates into K⁺ and Cl⁻ ions).
- K_f is the freezing point depression constant, which is a property of the solvent. For water, (K_f = 1.86^\circ C \, \text{m}^{-1}).
- m is the molality of the solution, which is calculated as moles of solute per kilogram of solvent.
Problem 1: 3.4 M NaCl Solution
- Calculate the molality (m):
- We are given the molarity (M) of the NaCl solution, which is 3.4 M. Molarity is defined as moles of solute per liter of solution. Since the density of the solution is given as 1.00 g/mL, we assume 1 L of solution weighs 1000 g.
- The mass of solvent (water) can be approximated as:
[
\text{Mass of solution} = 1000 \, \text{g}, \quad \text{Mass of NaCl} = 3.4 \, \text{mol} \times 58.44 \, \text{g/mol} = 198.7 \, \text{g}
]
[
\text{Mass of water (solvent)} = 1000 \, \text{g} – 198.7 \, \text{g} = 801.3 \, \text{g} = 0.8013 \, \text{kg}
] - Molality (m) is:
[
m = \frac{\text{moles of NaCl}}{\text{kg of solvent}} = \frac{3.4 \, \text{mol}}{0.8013 \, \text{kg}} \approx 4.24 \, \text{mol/kg}
]
- Calculate the freezing point depression:
Using the formula for freezing point depression:
[
\Delta T_f = i \cdot K_f \cdot m = 2 \cdot 1.86 \, \text{°C/m} \cdot 4.24 \, \text{mol/kg} \approx 15.8 \, \text{°C}
]
The freezing point depression for this solution is 15.8°C.
Problem 2: 27g of KCl in 1.6 L of Water
- Calculate the moles of KCl:
The molar mass of KCl is approximately 74.55 g/mol. Therefore, the number of moles of KCl in 27 g is:
[
\text{moles of KCl} = \frac{27 \, \text{g}}{74.55 \, \text{g/mol}} \approx 0.362 \, \text{mol}
] - Calculate the molality (m):
The volume of water is 1.6 L, and we assume the density of water is 1.00 g/mL, so the mass of water is:
[
\text{mass of water} = 1.6 \, \text{L} \times 1000 \, \text{g/L} = 1600 \, \text{g} = 1.6 \, \text{kg}
] Molality (m) is:
[
m = \frac{\text{moles of KCl}}{\text{kg of solvent}} = \frac{0.362 \, \text{mol}}{1.6 \, \text{kg}} \approx 0.226 \, \text{mol/kg}
] - Calculate the freezing point depression:
Using the formula for freezing point depression:
[
\Delta T_f = i \cdot K_f \cdot m = 2 \cdot 1.86 \, \text{°C/m} \cdot 0.226 \, \text{mol/kg} \approx 0.84 \, \text{°C}
]
The freezing point depression for this solution is 0.84°C.
Summary:
- For the 3.4 M NaCl solution, the freezing point depression is approximately 15.8°C.
- For the 27g KCl in 1.6 L of water solution, the freezing point depression is approximately 0.84°C.
These calculations show how the amount of solute, the solvent mass, and the dissociation factor (van’t Hoff factor) influence the freezing point depression. The greater the concentration of solute particles in the solution, the larger the depression of the freezing point.