Question 13.4.5: Calculate the boiling point of a solution. What is the boili......

You are asked to calculate the boiling point of a solution.
You are given the mass of solute and solvent in the solution.
The boiling point of the solution is the normal boiling point plus the boiling point elevation. Using data from Table 13.4.1,
T_{bp} = 80.1 °C + (2.53 °C/m)(m_{solute})i
The concentration of I_{2}, a nonelectrolyte, is

m_{I_{2}} = \frac{\text{mol I}_{2}}{\text{kg benzene}} = \frac{\left(40.0\text{ g}\right)\left(\frac{1{\text{ mol I}_{2}}}{253.8\text{ g}} \right) }{0.250 \text{ kg benzene}} = 0.630 m

Use equation 13.9 to calculate the boiling point of the solution

T_{bp} = 80.1 °C + (2.53 °C/m)(0.630 m)(1) = 81.7 °C

ΔT_{bp} = K_{bp}m_{solute}i         (13.9)

Is your answer reasonable? A solute was added to the solvent, so the boiling point of the solution is greater than the boiling point of the pure solvent.