Consider a solution containing a mole of toluene and a hundred moles of benzene. Kf=5.12°C/m Kb=2.65°C/m Melting point= 5.5°C Liquefaction point= 80.1°C What is the freezing point of the solution? A. 4.84°C B. 4.99°C C. 5.28°C D. 5.43°C
The Correct Answer and Explanation is:
Correct Answer: A. 4.84°C
Explanation:
To find the freezing point of the solution, we use the formula for freezing point depression: ΔTf=Kf×m\Delta T_f = K_f \times mΔTf=Kf×m
Where:
- ΔTf\Delta T_fΔTf is the freezing point depression
- KfK_fKf is the freezing point depression constant (for benzene, given as 5.12°C/m)
- mmm is the molality of the solute
Step 1: Identify known values
- 1 mole of toluene (solute)
- 100 moles of benzene (solvent)
- Freezing point of pure benzene is 5.5°C
- Kf=5.12K_f = 5.12Kf=5.12°C/m
We assume the mass of the solvent (benzene) is known or can be estimated. Since benzene has a molar mass of about 78 g/mol, 100 moles would be: 100 mol×78 g/mol=7800 g=7.8 kg100 \text{ mol} \times 78 \text{ g/mol} = 7800 \text{ g} = 7.8 \text{ kg}100 mol×78 g/mol=7800 g=7.8 kg
Step 2: Calculate molality of the solution
Molality (m) is moles of solute per kg of solvent: m=1 mol toluene7.8 kg benzene≈0.1282 mol/kgm = \frac{1 \text{ mol toluene}}{7.8 \text{ kg benzene}} \approx 0.1282 \text{ mol/kg}m=7.8 kg benzene1 mol toluene≈0.1282 mol/kg
Step 3: Calculate freezing point depression ΔTf=5.12×0.1282≈0.6566∘C\Delta T_f = 5.12 \times 0.1282 \approx 0.6566^\circ CΔTf=5.12×0.1282≈0.6566∘C
Step 4: Calculate new freezing point Freezing point=5.5∘C−0.6566∘C≈4.84∘C\text{Freezing point} = 5.5^\circ C – 0.6566^\circ C \approx 4.84^\circ CFreezing point=5.5∘C−0.6566∘C≈4.84∘C
Final Answer: A. 4.84°C
This result shows how adding a small amount of solute (toluene) lowers the freezing point of a solvent (benzene) due to the colligative property known as freezing point depression, which depends on the number of solute particles in the solution, not their identity.
