Carbon disulfide (CS2) boils at 46.30 °C and has a density of 1.261 g/mL.(a)When 0.250 mol of a nondissociating solute is dissolved in 400.0 mL of CS2, the solution boils at 47.46 °C . What is the molal boiling-point-elevation constant for CS2? (b) When 5.39 g of a nondissociating unknown is dissolved in 50.0 mL of CS2, the solution boils at 47.08 °C. What is the molecular weight of the unknown?
The Correct Answer and Explanation is :
To solve this, we will use the boiling-point elevation equation:
[
\Delta T_b = K_b \cdot m
]
where:
- (\Delta T_b) = boiling-point elevation,
- (K_b) = molal boiling-point elevation constant,
- (m) = molality of the solution.
(a) Determining (K_b) for CS(_2):
- Calculate (\Delta T_b):
[
\Delta T_b = 47.46^\circ \text{C} – 46.30^\circ \text{C} = 1.16^\circ \text{C}
] - Calculate the molality ((m)):
The solute is nondissociating, and its moles are given as (0.250). The mass of CS(_2) can be calculated using its density and volume:
[
\text{Mass of CS}_2 = 400.0 \, \text{mL} \times 1.261 \, \text{g/mL} = 504.4 \, \text{g}
]
Convert this to kilograms:
[
\text{Mass in kg} = \frac{504.4 \, \text{g}}{1000} = 0.5044 \, \text{kg}
]
Molality:
[
m = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{0.250}{0.5044} = 0.4957 \, \text{mol/kg}
] - Determine (K_b):
[
K_b = \frac{\Delta T_b}{m} = \frac{1.16}{0.4957} = 2.34 \, ^\circ \text{C·kg/mol}
]
(b) Determining the molecular weight of the unknown:
- Calculate (\Delta T_b):
[
\Delta T_b = 47.08^\circ \text{C} – 46.30^\circ \text{C} = 0.78^\circ \text{C}
] - Determine molality ((m)):
Rearrange the boiling-point elevation equation:
[
m = \frac{\Delta T_b}{K_b} = \frac{0.78}{2.34} = 0.3333 \, \text{mol/kg}
] - Calculate the mass of CS(_2):
[
\text{Mass of CS}_2 = 50.0 \, \text{mL} \times 1.261 \, \text{g/mL} = 63.05 \, \text{g}
]
Convert to kilograms:
[
\text{Mass in kg} = \frac{63.05 \, \text{g}}{1000} = 0.06305 \, \text{kg}
] - Determine moles of solute:
[
\text{Moles of solute} = m \times \text{kg of solvent} = 0.3333 \times 0.06305 = 0.02101 \, \text{mol}
] - Calculate molar mass (molecular weight):
The mass of the solute is (5.39 \, \text{g}), and its molar mass is:
[
\text{Molar mass} = \frac{\text{mass}}{\text{moles}} = \frac{5.39}{0.02101} = 256.5 \, \text{g/mol}
]
Final Answers:
- (a) (K_b = 2.34 \, ^\circ \text{C·kg/mol}),
- (b) Molecular weight = (256.5 \, \text{g/mol}).
Explanation:
The boiling-point elevation method is based on colligative properties, which depend only on the number of solute particles and not their identity. In (a), (K_b) is found using the known solute and solvent properties. In (b), the molecular weight is determined using the solute mass, boiling-point elevation, and solvent properties. These calculations demonstrate the interplay between boiling-point elevation, solute concentration, and molar mass.