Given:
mass of solute = 0.144 g
mass of solvent = 10.0 g = 0.0100 kg
molar mass of solute = 164.2 g/mol
[tex]\text{BP}_{solvent} = \text{80.10°C}[/tex]
[tex]K_{\text{b}} = \text{2.53°C/molal}[/tex]
solute: eugenol
solvent: benzene
Required:
[tex]\text{BP}_{solution}[/tex]
Solution:
Step 1: Calculate the number of moles of solute.
[tex]\text{moles of solute} = \frac{\text{mass of solute}}{\text{molar mass of solute}}[/tex]
[tex]\text{moles of solute} = \frac{\text{0.144 g}}{\text{164.2 g/mol}}[/tex]
[tex]\text{moles of solute} = 8.76979 × 10^{-4} \: \text{mol}[/tex]
Step 2: Calculate the molality of solution.
[tex]m = \frac{\text{moles of solute}}{\text{mass of solvent}}[/tex]
[tex]m = \frac{8.76979 × 10^{-4} \: \text{mol}}{\text{0.0100 kg}}[/tex]
[tex]m = \text{0.0876979 molal}[/tex]
Step 3: Calculate the boiling point elevation.
[tex]\Delta T_{\text{b}} = mK_{\text{b}}[/tex]
[tex]\Delta T_{\text{b}} = (\text{0.0876979 molal})(\text{2.53°C/molal})[/tex]
[tex]\Delta T_{\text{b}} = \text{0.222°C}[/tex]
Step 4: Calculate the boiling point of the solution.
[tex]\text{BP}_{solution} = \text{BP}_{solvent} + \Delta T_{\text{b}}[/tex]
[tex]\text{BP}_{solution} = \text{80.10°C} + \text{0.222°C}[/tex]
[tex]\boxed{\text{BP}_{solution} = \text{80.322°C}}[/tex]
[tex]\\[/tex]
#CarryOnLearning
