Benzene boils at 80.10 ° C and has a molal boiling constant, k_b, of 2.53 C/m. When 2.15 g of a compound is dissolved in 20.0 g of benzene, the resulting solution has a boiling point of 81.10 ° C. What is the molality of the solute?

Question

Quizplus · Accepted Answer

First, calculate the change in boiling point (ΔTb) using the formula:

ΔTb = Tb(solution) - Tb(solvent)
ΔTb = 81.10 °C - 80.10 °C = 1.00 °C

Next, calculate the molality (m) of the solute using the formula:

ΔTb = kbm
m = ΔTb / kb

where kb is the molal boiling constant for benzene (2.53 °C/m).

m = 1.00 °C / (2.53 °C/m)
m = 0.395 m

Therefore, the molality of the solute is 0.395 m, which corresponds to choice C.

Benzene Boils at 80