Step-by-Step Guidance for Colligative Properties Worksheet

Q1. Calculate the mass of urea that should be dissolved in 225 g of water at 35°C to produce a solution with a vapor pressure of 37.1 mmHg. (Molar mass: water = 18.02 g/mol, urea = 60.06 g/mol. Vapor pressure of water at 35°C = 42.2 mmHg.)

Background

Topic: Colligative Properties – Vapor Pressure Lowering (Raoult's Law)

This question tests your understanding of how a nonvolatile solute (urea) lowers the vapor pressure of a solvent (water) and how to use Raoult's Law to relate vapor pressure lowering to the amount of solute dissolved.

Key Terms and Formulas

• Raoult's Law: $P_{solution} = X_{solvent} \cdot P^0_{solvent}$

• $P_{solution}$ = vapor pressure of the solution

• $P^0_{solvent}$ = vapor pressure of pure solvent

• $X_{solvent}$ = mole fraction of the solvent

• Mole fraction: $X_{solvent} = \frac{n_{solvent}}{n_{solvent} + n_{solute}}$

Step-by-Step Guidance

1. Calculate the number of moles of water: $n_{water} = \frac{225\ \text{g}}{18.02\ \text{g/mol}}$

2. Let $n_{urea}$ be the number of moles of urea to be dissolved. The total moles in solution will be $n_{water} + n_{urea}$.

3. Write the expression for the mole fraction of water: $X_{water} = \frac{n_{water}}{n_{water} + n_{urea}}$

4. Apply Raoult's Law: $P_{solution} = X_{water} \cdot P^0_{water}$, where $P_{solution} = 37.1$ mmHg and $P^0_{water} = 42.2$ mmHg.

5. Rearrange to solve for $X_{water}$: $X_{water} = \frac{P_{solution}}{P^0_{water}}$

6. Set up the equation $\frac{n_{water}}{n_{water} + n_{urea}} = X_{water}$ and solve for $n_{urea}$.

7. Once you have $n_{urea}$, you can find the mass of urea using $\text{mass} = n_{urea} \times 60.06$ g/mol.

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Q2. What is the vapor pressure of a solution prepared by dissolving 0.75 moles of nonvolatile safrole in 950 g of ethanol at 25°C? (Molar mass of ethanol = 46.07 g/mol. Vapor pressure of ethanol at 25°C = 50.0 torr.)

Background

Topic: Colligative Properties – Vapor Pressure Lowering (Raoult's Law)

This question tests your ability to calculate the vapor pressure of a solution when a nonvolatile solute is dissolved in a volatile solvent, using Raoult's Law.

Key Terms and Formulas

• Raoult's Law: $P_{solution} = X_{solvent} \cdot P^0_{solvent}$

• $X_{solvent} = \frac{n_{solvent}}{n_{solvent} + n_{solute}}$

Step-by-Step Guidance

1. Calculate the number of moles of ethanol: $n_{ethanol} = \frac{950\ \text{g}}{46.07\ \text{g/mol}}$

2. Find the total moles in the solution: $n_{total} = n_{ethanol} + 0.75$

3. Calculate the mole fraction of ethanol: $X_{ethanol} = \frac{n_{ethanol}}{n_{total}}$

4. Apply Raoult's Law: $P_{solution} = X_{ethanol} \cdot 50.0$ torr

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Q3. When 0.0250 g of β-carotene is dissolved in 1.50 g of liquid camphor, the mixture has a freezing point of 178.6°C. What is the molar mass of β-carotene? (Freezing point constant for camphor = 40.0°C/m. Melting point of pure camphor = 179.8°C.)

Background

Topic: Colligative Properties – Freezing Point Depression

This question tests your ability to use freezing point depression to determine the molar mass of an unknown solute.

Key Terms and Formulas

• Freezing point depression: $\Delta T_f = K_f \cdot m$

• $\Delta T_f$ = change in freezing point ($T_{f,\text{pure}} - T_{f,\text{solution}}$)

• $K_f$ = freezing point depression constant (°C/m)

• $m$ = molality = $\frac{\text{moles solute}}{\text{kg solvent}}$

Step-by-Step Guidance

1. Calculate the freezing point depression: $\Delta T_f = 179.8 - 178.6$

2. Use $\Delta T_f = K_f \cdot m$ to solve for molality $m$.

3. Calculate the moles of β-carotene: $m = \frac{\text{moles β-carotene}}{1.50/1000}$ (since 1.50 g = 0.00150 kg)

4. Set up the equation to solve for moles of β-carotene.

5. Use the mass of β-carotene and the moles calculated to find the molar mass: $\text{Molar mass} = \frac{0.0250\ \text{g}}{\text{moles β-carotene}}$

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Q4. In each of the following lists of three solutions, circle the solution with the highest boiling point and underline the solution with the lowest boiling point. (Unless otherwise indicated, the following salts are strong electrolytes.)

Background

Topic: Colligative Properties – Boiling Point Elevation

This question tests your understanding of how the number of particles in solution (including ionization for electrolytes) affects boiling point elevation.

Key Terms and Formulas

• Boiling point elevation: $\Delta T_b = i K_b m$

• $i$ = van 't Hoff factor (number of particles the solute dissociates into)

• $K_b$ = boiling point elevation constant

• $m$ = molality

Step-by-Step Guidance

1. For each solution, determine the van 't Hoff factor $i$ (e.g., NaCl: $i=2$, Na2SO4: $i=3$, sucrose: $i=1$).

2. Calculate $i \times m$ for each solution to compare the effect on boiling point.

3. The solution with the highest $i \times m$ will have the highest boiling point; the lowest $i \times m$ will have the lowest boiling point.

4. Repeat this process for each set (a, b, c), considering whether the solute is a strong or weak electrolyte or a nonelectrolyte.

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Q5. What mass of insulin must be dissolved in water to make a 50.0 mL solution with an osmotic pressure of 16.8 mmHg at 25°C? (R = 0.08206 L·atm/(mol·K), 760 mmHg = 1 atm, molar mass of insulin = 5734 g/mol.)

Background

Topic: Colligative Properties – Osmotic Pressure

This question tests your ability to use the osmotic pressure equation to determine the mass of a solute needed to achieve a certain osmotic pressure.

Key Terms and Formulas

• Osmotic pressure: $\Pi = MRT$

• $\Pi$ = osmotic pressure (in atm)

• $M$ = molarity (mol/L)

• $R$ = gas constant (0.08206 L·atm/(mol·K))

• $T$ = temperature in Kelvin

Step-by-Step Guidance

1. Convert the osmotic pressure from mmHg to atm: $\Pi = \frac{16.8}{760}$ atm

2. Convert the temperature to Kelvin: $T = 25 + 273.15$

3. Rearrange the osmotic pressure equation to solve for molarity: $M = \frac{\Pi}{RT}$

4. Calculate the number of moles of insulin needed: $n = M \times 0.0500$ L

5. Find the mass of insulin: $\text{mass} = n \times 5734$ g/mol

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Q6. What mass of naphthalene must be dissolved in 200. g of benzene to give a solution with a freezing point 2.50°C below that of pure benzene? (Freezing point constant of benzene = 5.12°C/m, molar mass of naphthalene = 128.2 g/mol.)

Background

Topic: Colligative Properties – Freezing Point Depression

This question tests your ability to use freezing point depression to calculate the mass of a solute needed to achieve a specific freezing point lowering.

Key Terms and Formulas

• Freezing point depression: $\Delta T_f = K_f \cdot m$

• $m$ = molality = $\frac{\text{moles solute}}{\text{kg solvent}}$

Step-by-Step Guidance

1. Set $\Delta T_f = 2.50$°C and $K_f = 5.12$°C/m.

2. Calculate the molality: $m = \frac{\Delta T_f}{K_f}$

3. Convert 200. g of benzene to kg: $0.200$ kg

4. Calculate the moles of naphthalene needed: $m = \frac{n_{naph}}{0.200}$

5. Find the mass of naphthalene: $\text{mass} = n_{naph} \times 128.2$ g/mol

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Q7. Determine the boiling point and freezing point of a solution prepared by dissolving 678 g of glucose in 2.0 kg of water. (Boiling and freezing point constants for water: 0.52°C/m and 1.86°C/m, respectively. Molar mass of glucose = 180.2 g/mol. Water freezes at 0.0°C and boils at 100.0°C.)

Background

Topic: Colligative Properties – Boiling Point Elevation and Freezing Point Depression

This question tests your ability to calculate both the boiling point elevation and freezing point depression for a solution, and then determine the new boiling and freezing points.

Key Terms and Formulas

• Boiling point elevation: $\Delta T_b = i K_b m$

• Freezing point depression: $\Delta T_f = i K_f m$

• For glucose (a nonelectrolyte), $i = 1$

• Molality: $m = \frac{\text{moles solute}}{\text{kg solvent}}$

Step-by-Step Guidance

1. Calculate the moles of glucose: $n_{glucose} = \frac{678}{180.2}$

2. Calculate the molality: $m = \frac{n_{glucose}}{2.0}$

3. Calculate the boiling point elevation: $\Delta T_b = 1 \times 0.52 \times m$

4. Calculate the freezing point depression: $\Delta T_f = 1 \times 1.86 \times m$

5. Add $\Delta T_b$ to 100.0°C for the new boiling point; subtract $\Delta T_f$ from 0.0°C for the new freezing point.

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Q8. How many grams of sucrose (C12H22O11) must be added to 552 g of water to give a solution with a vapor pressure 2.00 mmHg less than that of pure water at 20°C? (Vapor pressure of water at 20°C = 17.5 mmHg. Molar mass of C12H22O11 = 342.3 g/mol, water = 18.02 g/mol.)

Background

Topic: Colligative Properties – Vapor Pressure Lowering (Raoult's Law)

This question tests your ability to use Raoult's Law to determine the amount of a nonvolatile solute needed to achieve a specific vapor pressure lowering.

Key Terms and Formulas

• Raoult's Law: $P_{solution} = X_{solvent} \cdot P^0_{solvent}$

• Vapor pressure lowering: $\Delta P = P^0_{solvent} - P_{solution}$

• Mole fraction: $X_{solvent} = \frac{n_{solvent}}{n_{solvent} + n_{solute}}$

Step-by-Step Guidance

1. Calculate the number of moles of water: $n_{water} = \frac{552}{18.02}$

2. Set $\Delta P = 2.00$ mmHg, so $P_{solution} = 17.5 - 2.00$ mmHg

3. Find $X_{water} = \frac{P_{solution}}{17.5}$

4. Set up the equation $X_{water} = \frac{n_{water}}{n_{water} + n_{sucrose}}$ and solve for $n_{sucrose}$

5. Calculate the mass of sucrose: $\text{mass} = n_{sucrose} \times 342.3$ g/mol

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