Determine the boiling point of a solution that contains 78.8 g of naphthalene (C10H8, molar mass = 128.16 g/mol) dissolved in 722 mL of benzene (d = 0.877 g/mL). Pure benzene has a boiling point of 80.1°C and a boiling point elevation constant of 2.53°C/m.
The Correct Answer and Explanation is :
The boiling point of the solution is approximately 82.56°C.
Explanation:
To determine the boiling point of a solution, we need to use the formula for boiling point elevation:
[
\Delta T_b = i \cdot K_b \cdot m
]
Where:
- (\Delta T_b) is the change in boiling point.
- (i) is the van ‘t Hoff factor, which represents the number of particles the solute dissociates into. Since naphthalene is a non-electrolyte and does not dissociate, (i = 1).
- (K_b) is the ebullioscopic constant (boiling point elevation constant) of the solvent (benzene), which is given as 2.53°C/m.
- (m) is the molality of the solution (moles of solute per kilogram of solvent).
Here are the detailed steps:
- Calculate the moles of naphthalene:
Using the mass of naphthalene and its molar mass:
[
\text{moles of naphthalene} = \frac{78.8 \, \text{g}}{128.16 \, \text{g/mol}} = 0.615 \, \text{mol}
] - Calculate the mass of benzene:
Using the density and volume of benzene:
[
\text{mass of benzene} = 722 \, \text{mL} \times 0.877 \, \text{g/mL} = 633.5 \, \text{g} = 0.6335 \, \text{kg}
] - Calculate the molality:
Molality (m) is moles of solute per kilogram of solvent:
[
m = \frac{0.615 \, \text{mol}}{0.6335 \, \text{kg}} = 0.970 \, \text{mol/kg}
] - Calculate the boiling point elevation:
Using the formula for boiling point elevation:
[
\Delta T_b = 2.53 \, \text{°C/m} \times 0.970 \, \text{mol/kg} = 2.45 \, \text{°C}
] - Calculate the new boiling point of the solution:
The new boiling point is the sum of the pure benzene boiling point and the elevation:
[
\text{New boiling point} = 80.1 \, \text{°C} + 2.45 \, \text{°C} = 82.56 \, \text{°C}
]
This method shows how the presence of a solute (naphthalene) raises the boiling point of the solvent (benzene).