Question

A 1.60-g sample of a mixture of naphthalene (C10H8) and anthracene (C14H10) is dissolved in 20.0 g benzene (C6H6). The freezing point of the solution is found to be 2.81 °C. What is the composition as mass percent of the sample mixture?

Answer 1

from the litreture

The freezing point of benzene is 5.51°C and Kf= 5.21°C*kg/mol

delta T = 5.51 - 2.81 = 2.7°C

2.7 = 5.21 x m

m = 0.518 = moles / 0.0200 Kg benzene

moles = 0.0104

let x = mass C10H8 (napthalene)
let y = mass C14H10 (anthracene)

x + y = 1.60
x = 1.60-y

x / 128.174 + y / 178.234 = 0.0104

1.60 -y / 128.174 + y / 178.234 = 0.0104

285.2 - 178.234 y + 128.174 y = 237.6

47.6 - 50.06 y = 0
y = (anthracene) = 0.951 g
x = (naphthalene) = 1.60 - 0.951 = 0.649 g