Colligative Properties Worksheet - Free Printable

Problem 1: Melting and Boiling Points of the Solution

#### Given:
- Mass of sodium chloride (NaCl): 45 grams
- Mass of water: 500 grams
- Molal freezing point depression constant for water, \( K_f(H_2O) = 0.52 \, ^\circ \text{C/m} \)
- Molal boiling point elevation constant for water, \( K_b(H_2O) = 0.86 \, ^\circ \text{C/m} \)

#### Objective:
Determine the melting point and boiling point of the resulting solution.

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#### Step 1: Calculate the molality of the solution
Molality (\( m \)) is defined as the number of moles of solute per kilogram of solvent.

1. Calculate the moles of NaCl:
- Molar mass of NaCl = \( 23 + 35.5 = 58.5 \, \text{g/mol} \)
- Moles of NaCl = \( \frac{\text{mass of NaCl}}{\text{molar mass of NaCl}} = \frac{45 \, \text{g}}{58.5 \, \text{g/mol}} \approx 0.77 \, \text{mol} \)

2. Calculate the mass of the solvent (water) in kilograms:
- Mass of water = 500 grams = 0.5 kg

3. Calculate the molality (\( m \)):
\[
m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}} = \frac{0.77 \, \text{mol}}{0.5 \, \text{kg}} = 1.54 \, \text{m}
\]

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#### Step 2: Calculate the freezing point depression
The formula for freezing point depression is:
\[
\Delta T_f = i \cdot K_f \cdot m
\]
where:
- \( i \) is the van't Hoff factor (for NaCl, \( i = 2 \) because it dissociates into \( \text{Na}^+ \) and \( \text{Cl}^- \))
- \( K_f \) is the freezing point depression constant
- \( m \) is the molality

Substitute the values:
\[
\Delta T_f = 2 \cdot 0.52 \, ^\circ \text{C/m} \cdot 1.54 \, \text{m} = 1.62 \, ^\circ \text{C}
\]

The freezing point of pure water is \( 0^\circ \text{C} \). Therefore, the new freezing point of the solution is:
\[
T_f = 0^\circ \text{C} - \Delta T_f = 0^\circ \text{C} - 1.62^\circ \text{C} = -1.62^\circ \text{C}
\]

---

#### Step 3: Calculate the boiling point elevation
The formula for boiling point elevation is:
\[
\Delta T_b = i \cdot K_b \cdot m
\]
where:
- \( i \) is the van't Hoff factor (for NaCl, \( i = 2 \))
- \( K_b \) is the boiling point elevation constant
- \( m \) is the molality

Substitute the values:
\[
\Delta T_b = 2 \cdot 0.86 \, ^\circ \text{C/m} \cdot 1.54 \, \text{m} = 2.64 \, ^\circ \text{C}
\]

The boiling point of pure water is \( 100^\circ \text{C} \). Therefore, the new boiling point of the solution is:
\[
T_b = 100^\circ \text{C} + \Delta T_b = 100^\circ \text{C} + 2.64^\circ \text{C} = 102.64^\circ \text{C}
\]

---

#### Final Answer for Problem 1:
\[
\boxed{-1.62^\circ \text{C}, 102.64^\circ \text{C}}
\]

---

Problem 2: Vapor Pressure of the Solution

#### Given:
- Vapor pressure of pure water at \( 25^\circ \text{C} \): \( P^\circ = 3.17 \, \text{kPa} \)
- Solution from Problem 1: 45 grams of NaCl in 500 grams of water
- Molality of the solution: \( m = 1.54 \, \text{m} \)
- Van't Hoff factor for NaCl: \( i = 2 \)

#### Objective:
Determine the vapor pressure of the solution at \( 25^\circ \text{C} \).

---

#### Step 1: Use Raoult's Law
Raoult's Law states:
\[
P = P^\circ \cdot X_{\text{solvent}}
\]
where:
- \( P \) is the vapor pressure of the solution
- \( P^\circ \) is the vapor pressure of the pure solvent
- \( X_{\text{solvent}} \) is the mole fraction of the solvent

The mole fraction of the solvent can be calculated using:
\[
X_{\text{solvent}} = \frac{\text{moles of solvent}}{\text{total moles of solution}}
\]

1. Calculate the moles of water:
- Molar mass of water = \( 18 \, \text{g/mol} \)
- Moles of water = \( \frac{\text{mass of water}}{\text{molar mass of water}} = \frac{500 \, \text{g}}{18 \, \text{g/mol}} \approx 27.78 \, \text{mol} \)

2. Calculate the total moles of the solution:
- Moles of solute (NaCl) = 0.77 mol (from Problem 1)
- Total moles = moles of water + moles of solute = \( 27.78 + 0.77 = 28.55 \, \text{mol} \)

3. Calculate the mole fraction of water:
\[
X_{\text{water}} = \frac{\text{moles of water}}{\text{total moles}} = \frac{27.78}{28.55} \approx 0.973
\]

4. Calculate the vapor pressure of the solution:
\[
P = P^\circ \cdot X_{\text{water}} = 3.17 \, \text{kPa} \cdot 0.973 \approx 3.08 \, \text{kPa}
\]

---

#### Final Answer for Problem 2:
\[
\boxed{3.08 \, \text{kPa}}
\]

---

Problem 3: Comparing Boiling Points of Two Solutions

#### Given:
1. Solution A: 105 grams of sucrose (\( C_{12}H_{22}O_{11} \)) in 500 grams of water
2. Solution B: 35 grams of sodium chloride (NaCl) in 500 grams of water

#### Objective:
Determine which solution has a higher boiling point.

---

#### Step 1: Calculate the molality of Solution A (sucrose)
1. Calculate the moles of sucrose:
- Molar mass of sucrose = \( 12 \times 12 + 22 \times 1 + 11 \times 16 = 342 \, \text{g/mol} \)
- Moles of sucrose = \( \frac{\text{mass of sucrose}}{\text{molar mass of sucrose}} = \frac{105 \, \text{g}}{342 \, \text{g/mol}} \approx 0.307 \, \text{mol} \)

2. Calculate the molality (\( m \)):
- Mass of water = 500 grams = 0.5 kg
\[
m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}} = \frac{0.307 \, \text{mol}}{0.5 \, \text{kg}} = 0.614 \, \text{m}
\]

3. Calculate the boiling point elevation for Solution A:
- For sucrose, \( i = 1 \) (it does not dissociate)
\[
\Delta T_b = i \cdot K_b \cdot m = 1 \cdot 0.86 \, ^\circ \text{C/m} \cdot 0.614 \, \text{m} \approx 0.53 \, ^\circ \text{C}
\]

4. Calculate the boiling point of Solution A:
\[
T_b = 100^\circ \text{C} + \Delta T_b = 100^\circ \text{C} + 0.53^\circ \text{C} = 100.53^\circ \text{C}
\]

---

#### Step 2: Calculate the molality of Solution B (NaCl)
1. Calculate the moles of NaCl:
- Molar mass of NaCl = 58.5 g/mol
- Moles of NaCl = \( \frac{\text{mass of NaCl}}{\text{molar mass of NaCl}} = \frac{35 \, \text{g}}{58.5 \, \text{g/mol}} \approx 0.60 \, \text{mol} \)

2. Calculate the molality (\( m \)):
- Mass of water = 500 grams = 0.5 kg
\[
m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}} = \frac{0.60 \, \text{mol}}{0.5 \, \text{kg}} = 1.20 \, \text{m}
\]

3. Calculate the boiling point elevation for Solution B:
- For NaCl, \( i = 2 \)
\[
\Delta T_b = i \cdot K_b \cdot m = 2 \cdot 0.86 \, ^\circ \text{C/m} \cdot 1.20 \, \text{m} \approx 2.06 \, ^\circ \text{C}
\]

4. Calculate the boiling point of Solution B:
\[
T_b = 100^\circ \text{C} + \Delta T_b = 100^\circ \text{C} + 2.06^\circ \text{C} = 102.06^\circ \text{C}
\]

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#### Step 3: Compare the boiling points
- Boiling point of Solution A: \( 100.53^\circ \text{C} \)
- Boiling point of Solution B: \( 102.06^\circ \text{C} \)

Solution B has a higher boiling point.

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#### Final Answer for Problem 3:
\[
\boxed{\text{Solution B}}
\]