Question

A .435 g sample of a nonvolatile, nonionizable solute with a molar mass of 67.4 g/mol is dissolved in 15.1 g of cyclohexane. Calculate the molar concentration of solute in the solution.

Calculate the boiling point of the solution described.

Calculate the freezing point of the solution described.

Answer 1

we know that

moles = mass / molar mass

so

moles of solute = 0.435 / 67.4

moles of solute = 6.454 x 10-3

now

we know that

volume = mass / density

and

density of cyclohexane = 0.8 g/ml

so

volume of cyclohexane = 15.1 / 0.8

volume of cyclohexane = 18.875

assuming the volume of cyclohexane as the volume of soltuion

we get

volume of solution = 18.875 ml

now

molarity = moles of solute / volume of solution (L)

= 6.454 x 10-3 / 18.875 x 10-3

= 0.342

so

the molar conc is 0.342 M

now

we know that

molality = moles of solute / mass of solvent (kg)

so

molality = 6.454 x 10-3 / 15.1 x 10-3

molality = .4274

now

1)

elevation in boiling point is given by

dTb = kb x m

T - Tb = kb x m

for

cyclohexane

kb = 2.79

Tb = 80.74

so

T - 80.74 = 2.79 x 0.4274

T = 81.93

so

the boiling point of the solution is 81.93 C

2)

depression in freezing point is given by

dTf = kf x m

T - Tf = kf x m

for

cyclohexane

kf = -20.2

Tf = 6.55

so

T - 6.55 = -20.2 x 0.4274

T = -2.08

so

the freezing point of the solution is -2.08 C