200 grams of MgCl2 are added to 1 L of liquid water at standard conditions. a) Calculate the temperature at which water will freeze at 1 atm. b) Calculate the temperature at which water will boil at 1 atm
200 grams of MgCl2 are added to 1 L of liquid water at standard conditions.
a) Calculate the temperature at which water will freeze at 1 atm.
b) Calculate the temperature at which water will boil at 1 atm
The correct answer and explanation is :
To solve this problem, we will use the freezing point depression and boiling point elevation formulas.
Step 1: Determine the number of moles of MgCl₂
The molar mass of MgCl₂:
[
\text{Mg} = 24.305 \text{ g/mol}, \quad \text{Cl} = 2 \times 35.453 \text{ g/mol}
]
[
= 24.305 + 70.906 = 95.211 \text{ g/mol}
]
Moles of MgCl₂ in 200 g:
[
\frac{200 \text{ g}}{95.211 \text{ g/mol}} = 2.1 \text{ moles}
]
Since MgCl₂ dissociates into Mg²⁺ and 2 Cl⁻, the number of particles produced:
[
\text{van’t Hoff factor}, i = 3
]
Total moles of particles:
[
2.1 \times 3 = 6.3 \text{ moles of particles}
]
Step 2: Freezing Point Depression
The formula is:
[
\Delta T_f = i K_f m
]
where:
- ( K_f ) = 1.86 °C·kg/mol (freezing point depression constant for water)
- ( m ) = molality = (moles of particles) / (kg of solvent)
Water mass = 1 kg (since 1 L ≈ 1 kg for water)
[
m = \frac{6.3}{1} = 6.3 \text{ m}
]
[
\Delta T_f = 3 \times 1.86 \times 6.3 = 35.1^\circ \text{C}
]
New freezing point:
[
T_f = 0^\circ C – 35.1^\circ C = -35.1^\circ C
]
Step 3: Boiling Point Elevation
The formula is:
[
\Delta T_b = i K_b m
]
where:
- ( K_b ) = 0.512 °C·kg/mol (boiling point elevation constant for water)
[
\Delta T_b = 3 \times 0.512 \times 6.3 = 9.67^\circ \text{C}
]
New boiling point:
[
T_b = 100^\circ C + 9.67^\circ C = 109.67^\circ C
]
Final Answer:
a) Freezing Point = -35.1°C
b) Boiling Point = 109.67°C
The image visually represents the change in freezing and boiling points due to the addition of MgCl₂. You can see that the freezing point drops significantly to -35.1°C, while the boiling point rises to 109.67°C.