Question

Assuming ideal solution behavior, what is the boiling point of a solution of 9.04 g of I2 in 75.5 g of benzene, assuming the I2 is nonvolatile?

a) 80.1∘C
b) 91.7∘C
c) 86.4∘C
d) 98.2∘C

Answer 1

To determine the boiling point elevation of a solution, one must calculate the molality, multiply it by the solvent's ebullioscopic constant, and then add the result to the solvent's normal boiling point. Upon calculating, the boiling point of the I2 in benzene is approximately 80.2 °C; thus, the closest answer is 80.1 °C.

Step-by-step explanation:

To find the boiling point of the solution of I2 in benzene, we need to calculate the molality of the solution and then apply the formula for boiling point elevation.

The formula for boiling point elevation is:

ΔT

b

= K

b

m

where ΔTb is the boiling point elevation, Kb is the ebullioscopic constant of benzene, and m is the molality of the solution.

First, we calculate the molality of the solution:

m = moles of solute / kilograms of solvent

To find the moles of I2, we use:

moles of I2 = mass of I2 / molecular weight of I2 = 9.04 g / 253.8 g/mol

Next, we calculate the molality:

m = (9.04 g / 253.8 g/mol) / (75.5 g / 1000 g/kg) = 0.0473 mol/kg

Using the ebullioscopic constant of benzene (Kb = 2.53 °C/m), we can now calculate the boiling point elevation:

ΔTb = (2.53 °C/m)(0.0473 m) = 0.1196 °C

The normal boiling point of benzene is 80.1 °C, so the boiling point of the solution is:

boiling point = 80.1 °C + 0.1196 °C ≈ 80.2 °C

Therefore, the correct answer is 80.1 °C, which is closest to our calculated value, making (a) the best choice.