Answer :
To determine the freezing point of a solution containing 225 grams of calcium chloride (CaCl₂) in 500 grams of water, we'll employ the concept of freezing point depression. Let’s proceed step-by-step:
### Step 1: Calculate the Molar Mass of CaCl₂
First, we need to determine the molar mass of CaCl₂. The atomic masses are approximately:
- Calcium (Ca): 40.08 g/mol
- Chlorine (Cl): 35.453 g/mol
Therefore, the molar mass of CaCl₂ is:
[tex]\[ M_{\text{CaCl₂}} = 40.08 + (2 \times 35.453) = 40.08 + 70.906 = 110.986 \text{ g/mol} \][/tex]
### Step 2: Calculate the Number of Moles of CaCl₂
Next, we find the moles of CaCl₂ using the mass given:
[tex]\[ \text{moles of CaCl₂} = \frac{\text{mass}}{\text{molar mass}} = \frac{225 \text{ g}}{110.986 \text{ g/mol}} \approx 2.03 \text{ moles} \][/tex]
### Step 3: Determine the Van't Hoff Factor (i)
Calcium chloride (CaCl₂) dissociates in water into 3 ions:
- 1 Ca²⁺
- 2 Cl⁻
Thus, the Van't Hoff factor [tex]\( i \)[/tex] is 3.
### Step 4: Calculate the Molality of the Solution
Molality (m) is defined as the moles of solute per kilogram of solvent (water in this case):
[tex]\[ \text{mass of water} = 500 \text{ g} = 0.500 \text{ kg} \][/tex]
[tex]\[ \text{molality} (m) = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{2.03 \text{ moles}}{0.500 \text{ kg}} = 4.06 \text{ m} \][/tex]
### Step 5: Use the Freezing Point Depression Formula
The freezing point depression ([tex]\(\Delta T_f\)[/tex]) is given by:
[tex]\[ \Delta T_f = i \cdot K_f \cdot m \][/tex]
Where:
- [tex]\( i = 3 \)[/tex]
- [tex]\( K_f \)[/tex] is the cryoscopic constant for water, which is [tex]\( 1.86 \, \text{°C/m} \)[/tex]
- [tex]\( m = 4.06 \, \text{m} \)[/tex]
Substituting the values:
[tex]\[ \Delta T_f = 3 \times 1.86 \times 4.06 \][/tex]
[tex]\[ \Delta T_f = 3 \times 7.5516 \][/tex]
[tex]\[ \Delta T_f = 22.6548 \, \text{°C} \][/tex]
### Step 6: Calculate the New Freezing Point of the Solution
The normal freezing point of pure water is [tex]\( 0 \, °C \)[/tex]. The freezing point of the solution will be lowered by [tex]\(\Delta T_f\)[/tex]:
[tex]\[ \text{Freezing point of the solution} = 0 \, \text{°C} - 22.6548 \, \text{°C} \approx -22.65 \, \text{°C} \][/tex]
### Final Answer
The freezing point of the solution containing 225 g of CaCl₂ in 500 g of water is approximately [tex]\(-22.65 \, \text{°C}\)[/tex].
### Step 1: Calculate the Molar Mass of CaCl₂
First, we need to determine the molar mass of CaCl₂. The atomic masses are approximately:
- Calcium (Ca): 40.08 g/mol
- Chlorine (Cl): 35.453 g/mol
Therefore, the molar mass of CaCl₂ is:
[tex]\[ M_{\text{CaCl₂}} = 40.08 + (2 \times 35.453) = 40.08 + 70.906 = 110.986 \text{ g/mol} \][/tex]
### Step 2: Calculate the Number of Moles of CaCl₂
Next, we find the moles of CaCl₂ using the mass given:
[tex]\[ \text{moles of CaCl₂} = \frac{\text{mass}}{\text{molar mass}} = \frac{225 \text{ g}}{110.986 \text{ g/mol}} \approx 2.03 \text{ moles} \][/tex]
### Step 3: Determine the Van't Hoff Factor (i)
Calcium chloride (CaCl₂) dissociates in water into 3 ions:
- 1 Ca²⁺
- 2 Cl⁻
Thus, the Van't Hoff factor [tex]\( i \)[/tex] is 3.
### Step 4: Calculate the Molality of the Solution
Molality (m) is defined as the moles of solute per kilogram of solvent (water in this case):
[tex]\[ \text{mass of water} = 500 \text{ g} = 0.500 \text{ kg} \][/tex]
[tex]\[ \text{molality} (m) = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{2.03 \text{ moles}}{0.500 \text{ kg}} = 4.06 \text{ m} \][/tex]
### Step 5: Use the Freezing Point Depression Formula
The freezing point depression ([tex]\(\Delta T_f\)[/tex]) is given by:
[tex]\[ \Delta T_f = i \cdot K_f \cdot m \][/tex]
Where:
- [tex]\( i = 3 \)[/tex]
- [tex]\( K_f \)[/tex] is the cryoscopic constant for water, which is [tex]\( 1.86 \, \text{°C/m} \)[/tex]
- [tex]\( m = 4.06 \, \text{m} \)[/tex]
Substituting the values:
[tex]\[ \Delta T_f = 3 \times 1.86 \times 4.06 \][/tex]
[tex]\[ \Delta T_f = 3 \times 7.5516 \][/tex]
[tex]\[ \Delta T_f = 22.6548 \, \text{°C} \][/tex]
### Step 6: Calculate the New Freezing Point of the Solution
The normal freezing point of pure water is [tex]\( 0 \, °C \)[/tex]. The freezing point of the solution will be lowered by [tex]\(\Delta T_f\)[/tex]:
[tex]\[ \text{Freezing point of the solution} = 0 \, \text{°C} - 22.6548 \, \text{°C} \approx -22.65 \, \text{°C} \][/tex]
### Final Answer
The freezing point of the solution containing 225 g of CaCl₂ in 500 g of water is approximately [tex]\(-22.65 \, \text{°C}\)[/tex].
