Solved CHEMISTRY COLLIGATIVE PROPERTIES WORKSHEET Practice | Chegg.com - Free Printable

Problem Analysis:

The worksheet involves calculations related to colligative properties of solutions, specifically focusing on molality, freezing point depression, and boiling point elevation. Let's break down the problems step by step.

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Part A: Calculate the molality, freezing point, and boiling point for each water solution of nonionizing solutes

#### Formulae Needed:
1. Molality (m):
\[
m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}
\]
where:
- Moles of solute = \(\frac{\text{mass of solute}}{\text{molar mass of solute}}\)
- Solvent is given in kilograms.

2. Freezing Point Depression (\(\Delta T_f\)):
\[
\Delta T_f = K_f \cdot m
\]
where:
- \(K_f\) for water = 1.86 °C·kg/mol (freezing point depression constant for water).

3. Boiling Point Elevation (\(\Delta T_b\)):
\[
\Delta T_b = K_b \cdot m
\]
where:
- \(K_b\) for water = 0.512 °C·kg/mol (boiling point elevation constant for water).

4. New Freezing Point:
\[
T_f = T_f^o - \Delta T_f
\]
where:
- \(T_f^o\) = 0 °C (normal freezing point of water).

5. New Boiling Point:
\[
T_b = T_b^o + \Delta T_b
\]
where:
- \(T_b^o\) = 100 °C (normal boiling point of water).

---

Solutions for Part A:

#### Problem 1: 144 g of C₆H₁₂O₆ dissolved in 1000 g of H₂O
- Step 1: Calculate moles of C₆H₁₂O₆ (solute).
- Molar mass of C₆H₁₂O₆ = \(6 \times 12 + 12 \times 1 + 6 \times 16 = 72 + 12 + 96 = 180 \, \text{g/mol}\).
- Moles of C₆H₁₂O₆ = \(\frac{144 \, \text{g}}{180 \, \text{g/mol}} = 0.8 \, \text{mol}\).

- Step 2: Calculate molality (\(m\)).
- Mass of solvent (H₂O) = 1000 g = 1 kg.
- Molality (\(m\)) = \(\frac{0.8 \, \text{mol}}{1 \, \text{kg}} = 0.8 \, \text{m}\).

- Step 3: Calculate freezing point depression (\(\Delta T_f\)).
- \(\Delta T_f = K_f \cdot m = 1.86 \, \text{°C·kg/mol} \times 0.8 \, \text{m} = 1.488 \, \text{°C}\).

- Step 4: Calculate new freezing point.
- \(T_f = T_f^o - \Delta T_f = 0 \, \text{°C} - 1.488 \, \text{°C} = -1.488 \, \text{°C}\).

- Step 5: Calculate boiling point elevation (\(\Delta T_b\)).
- \(\Delta T_b = K_b \cdot m = 0.512 \, \text{°C·kg/mol} \times 0.8 \, \text{m} = 0.4096 \, \text{°C}\).

- Step 6: Calculate new boiling point.
- \(T_b = T_b^o + \Delta T_b = 100 \, \text{°C} + 0.4096 \, \text{°C} = 100.4096 \, \text{°C}\).

Final Answer for Problem 1:
\[
\boxed{0.8 \, \text{m}, -1.488 \, \text{°C}, 100.4096 \, \text{°C}}
\]

---

#### Problem 2: 48 g of CH₃OH dissolved in 200 g of H₂O
- Step 1: Calculate moles of CH₃OH (solute).
- Molar mass of CH₃OH = \(12 + 4 \times 1 + 16 = 32 \, \text{g/mol}\).
- Moles of CH₃OH = \(\frac{48 \, \text{g}}{32 \, \text{g/mol}} = 1.5 \, \text{mol}\).

- Step 2: Calculate molality (\(m\)).
- Mass of solvent (H₂O) = 200 g = 0.2 kg.
- Molality (\(m\)) = \(\frac{1.5 \, \text{mol}}{0.2 \, \text{kg}} = 7.5 \, \text{m}\).

- Step 3: Calculate freezing point depression (\(\Delta T_f\)).
- \(\Delta T_f = K_f \cdot m = 1.86 \, \text{°C·kg/mol} \times 7.5 \, \text{m} = 13.95 \, \text{°C}\).

- Step 4: Calculate new freezing point.
- \(T_f = T_f^o - \Delta T_f = 0 \, \text{°C} - 13.95 \, \text{°C} = -13.95 \, \text{°C}\).

- Step 5: Calculate boiling point elevation (\(\Delta T_b\)).
- \(\Delta T_b = K_b \cdot m = 0.512 \, \text{°C·kg/mol} \times 7.5 \, \text{m} = 3.84 \, \text{°C}\).

- Step 6: Calculate new boiling point.
- \(T_b = T_b^o + \Delta T_b = 100 \, \text{°C} + 3.84 \, \text{°C} = 103.84 \, \text{°C}\).

Final Answer for Problem 2:
\[
\boxed{7.5 \, \text{m}, -13.95 \, \text{°C}, 103.84 \, \text{°C}}
\]

---

#### Problem 3: 184 g of C₂H₅OH dissolved in 400 g of H₂O
- Step 1: Calculate moles of C₂H₅OH (solute).
- Molar mass of C₂H₅OH = \(2 \times 12 + 6 \times 1 + 16 = 24 + 6 + 16 = 46 \, \text{g/mol}\).
- Moles of C₂H₅OH = \(\frac{184 \, \text{g}}{46 \, \text{g/mol}} = 4 \, \text{mol}\).

- Step 2: Calculate molality (\(m\)).
- Mass of solvent (H₂O) = 400 g = 0.4 kg.
- Molality (\(m\)) = \(\frac{4 \, \text{mol}}{0.4 \, \text{kg}} = 10 \, \text{m}\).

- Step 3: Calculate freezing point depression (\(\Delta T_f\)).
- \(\Delta T_f = K_f \cdot m = 1.86 \, \text{°C·kg/mol} \times 10 \, \text{m} = 18.6 \, \text{°C}\).

- Step 4: Calculate new freezing point.
- \(T_f = T_f^o - \Delta T_f = 0 \, \text{°C} - 18.6 \, \text{°C} = -18.6 \, \text{°C}\).

- Step 5: Calculate boiling point elevation (\(\Delta T_b\)).
- \(\Delta T_b = K_b \cdot m = 0.512 \, \text{°C·kg/mol} \times 10 \, \text{m} = 5.12 \, \text{°C}\).

- Step 6: Calculate new boiling point.
- \(T_b = T_b^o + \Delta T_b = 100 \, \text{°C} + 5.12 \, \text{°C} = 105.12 \, \text{°C}\).

Final Answer for Problem 3:
\[
\boxed{10 \, \text{m}, -18.6 \, \text{°C}, 105.12 \, \text{°C}}
\]

---

#### Problem 4: 600 g of C₃H₇OH dissolved in 600 g of H₂O
- Step 1: Calculate moles of C₃H₇OH (solute).
- Molar mass of C₃H₇OH = \(3 \times 12 + 8 \times 1 + 16 = 36 + 8 + 16 = 60 \, \text{g/mol}\).
- Moles of C₃H₇OH = \(\frac{600 \, \text{g}}{60 \, \text{g/mol}} = 10 \, \text{mol}\).

- Step 2: Calculate molality (\(m\)).
- Mass of solvent (H₂O) = 600 g = 0.6 kg.
- Molality (\(m\)) = \(\frac{10 \, \text{mol}}{0.6 \, \text{kg}} = 16.67 \, \text{m}\).

- Step 3: Calculate freezing point depression (\(\Delta T_f\)).
- \(\Delta T_f = K_f \cdot m = 1.86 \, \text{°C·kg/mol} \times 16.67 \, \text{m} = 31.0022 \, \text{°C}\).

- Step 4: Calculate new freezing point.
- \(T_f = T_f^o - \Delta T_f = 0 \, \text{°C} - 31.0022 \, \text{°C} = -31.0022 \, \text{°C}\).

- Step 5: Calculate boiling point elevation (\(\Delta T_b\)).
- \(\Delta T_b = K_b \cdot m = 0.512 \, \text{°C·kg/mol} \times 16.67 \, \text{m} = 8.533 \, \text{°C}\).

- Step 6: Calculate new boiling point.
- \(T_b = T_b^o + \Delta T_b = 100 \, \text{°C} + 8.533 \, \text{°C} = 108.533 \, \text{°C}\).

Final Answer for Problem 4:
\[
\boxed{16.67 \, \text{m}, -31.0022 \, \text{°C}, 108.533 \, \text{°C}}
\]

---

#### Problem 5: 100 g of C₂H₆O₂ dissolved in 200 g of H₂O
- Step 1: Calculate moles of C₂H₆O₂ (solute).
- Molar mass of C₂H₆O₂ = \(2 \times 12 + 6 \times 1 + 2 \times 16 = 24 + 6 + 32 = 62 \, \text{g/mol}\).
- Moles of C₂H₆O₂ = \(\frac{100 \, \text{g}}{62 \, \text{g/mol}} = 1.6129 \, \text{mol}\).

- Step 2: Calculate molality (\(m\)).
- Mass of solvent (H₂O) = 200 g = 0.2 kg.
- Molality (\(m\)) = \(\frac{1.6129 \, \text{mol}}{0.2 \, \text{kg}} = 8.0645 \, \text{m}\).

- Step 3: Calculate freezing point depression (\(\Delta T_f\)).
- \(\Delta T_f = K_f \cdot m = 1.86 \, \text{°C·kg/mol} \times 8.0645 \, \text{m} = 15.05 \, \text{°C}\).

- Step 4: Calculate new freezing point.
- \(T_f = T_f^o - \Delta T_f = 0 \, \text{°C} - 15.05 \, \text{°C} = -15.05 \, \text{°C}\).

- Step 5: Calculate boiling point elevation (\(\Delta T_b\)).
- \(\Delta T_b = K_b \cdot m = 0.512 \, \text{°C·kg/mol} \times 8.0645 \, \text{m} = 4.13 \, \text{°C}\).

- Step 6: Calculate new boiling point.
- \(T_b = T_b^o + \Delta T_b = 100 \, \text{°C} + 4.13 \, \text{°C} = 104.13 \, \text{°C}\).

Final Answer for Problem 5:
\[
\boxed{8.06 \, \text{m}, -15.05 \, \text{°C}, 104.13 \, \text{°C}}
\]

---

Part B: Calculate the molality of a water solution if the freezing point is:

#### Formula Used:
\[
\Delta T_f = K_f \cdot m \quad \Rightarrow \quad m = \frac{\Delta T_f}{K_f}
\]
where:
- \(\Delta T_f = T_f^o - T_f\)
- \(K_f = 1.86 \, \text{°C·kg/mol}\)

---

#### Problem 6: Freezing point = -9.3°C
- Step 1: Calculate \(\Delta T_f\).
- \(\Delta T_f = T_f^o - T_f = 0 \, \text{°C} - (-9.3 \, \text{°C}) = 9.3 \, \text{°C}\).

- Step 2: Calculate molality (\(m\)).
- \(m = \frac{\Delta T_f}{K_f} = \frac{9.3 \, \text{°C}}{1.86 \, \text{°C·kg/mol}} = 5 \, \text{m}\).

Final Answer for Problem 6:
\[
\boxed{5 \, \text{m}}
\]

---

#### Problem 7: Freezing point = -27.9°C
- Step 1: Calculate \(\Delta T_f\).
- \(\Delta T_f = T_f^o - T_f = 0 \, \text{°C} - (-27.9 \, \text{°C}) = 27.9 \, \text{°C}\).

- Step 2: Calculate molality (\(m\)).
- \(m = \frac{\Delta T_f}{K_f} = \frac{27.9 \, \text{°C}}{1.86 \, \text{°C·kg/mol}} = 15 \, \text{m}\).

Final Answer for Problem 7:
\[
\boxed{15 \, \text{m}}
\]

---

#### Problem 8: Freezing point = -7.44°C
- Step 1: Calculate \(\Delta T_f\).
- \(\Delta T_f = T_f^o - T_f = 0 \, \text{°C} - (-7.44 \, \text{°C}) = 7.44 \, \text{°C}\).

- Step 2: Calculate molality (\(m\)).
- \(m = \frac{\Delta T_f}{K_f} = \frac{7.44 \, \text{°C}}{1.86 \, \text{°C·kg/mol}} = 4 \, \text{m}\).

Final Answer for Problem 8:
\[
\boxed{4 \, \text{m}}
\]

---

Part C: Calculate the molality of a water solution if the boiling point is:

#### Formula Used:
\[
\Delta T_b = K_b \cdot m \quad \Rightarrow \quad m = \frac{\Delta T_b}{K_b}
\]
where:
- \(\Delta T_b = T_b - T_b^o\)
- \(K_b = 0.512 \, \text{°C·kg/mol}\)

---

#### Problem 9: Boiling point = 103.12°C
- Step 1: Calculate \(\Delta T_b\).
- \(\Delta T_b = T_b - T_b^o = 103.12 \, \text{°C} - 100 \, \text{°C} = 3.12 \, \text{°C}\).

- Step 2: Calculate molality (\(m\)).
- \(m = \frac{\Delta T_b}{K_b} = \frac{3.12 \, \text{°C}}{0.512 \, \text{°C·kg/mol}} = 6.09 \, \text{m}\).

Final Answer for Problem 9:
\[
\boxed{6.09 \, \text{m}}
\]

---

#### Problem 10: Boiling point = 108.32°C
- Step 1: Calculate \(\Delta T_b\).
- \(\Delta T_b = T_b - T_b^o = 108.32 \, \text{°C} - 100 \, \text{°C} = 8.32 \, \text{°C}\).

- Step 2: Calculate molality (\(m\)).
- \(m = \frac{\Delta T_b}{K_b} = \frac{8.32 \, \text{°C}}{0.512 \, \text{°C·kg/mol}} = 16.25 \, \text{m}\).

Final Answer for Problem 10:
\[
\boxed{16.25 \, \text{m}}
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
&\text{Part A:} \\
&1. \, 0.8 \, \text{m}, -1.488 \, \text{°C}, 100.4096 \, \text{°C} \\
&2. \, 7.5 \, \text{m}, -13.95 \, \text{°C}, 103.84 \, \text{°C} \\
&3. \, 10 \, \text{m}, -18.6 \, \text{°C}, 105.12 \, \text{°C} \\
&4. \, 16.67 \, \text{m}, -31.0022 \, \text{°C}, 108.533 \, \text{°C} \\
&5. \, 8.06 \, \text{m}, -15.05 \, \text{°C}, 104.13 \, \text{°C} \\
\\
&\text{Part B:} \\
&6. \, 5 \, \text{m} \\
&7. \, 15 \, \text{m} \\
&8. \, 4 \, \text{m} \\
\\
&\text{Part C:} \\
&9. \, 6.09 \, \text{m} \\
&10. \, 16.25 \, \text{m}
\end{aligned}
}
\]