Question

John performs a similar colligative property experiment as Exp 22. He prepared a solution by mixing 11.2 g of naphthalene, C₁₀H₈(s), in 350.0 g of cyclohexane.If John measured the freezing point of this solution to be 1.28 celsius, what is the relative error in John's experiment?

	Molar mass	Freezing Point	Freezing Point Depression
Cyclohexane	84.16 g/mol	6.55 celsius	20.20 celsius/m
Napthalene	128.17 g/mol	80.26 celsius	6.80 celsius/m

Answer 1

Apply Colligative properties

This is a typical example of colligative properties.

Recall that a solute ( non volatile ) can make a depression/increase in the freezing/boiling point via:

dTf = -Kf*molality * i

dTb = Kb*molality * i

where:

Kf = freezing point constant for the SOLVENT; Kb = boiling point constant for the SOLVENT;

molality = moles of SOLUTE / kg of SOLVENT

i = vant hoff coefficient, typically the total ion/molecular concentration.

At the end:

Tf mix = Tf solvent - dTf

substitute data

1.28 = 6.55 -Kf*molality * i

1.28 = 6.55 -(20.20) *mass/MW / kg solvent

m = 11.2 g of N

m = 350 g of CH = 0.350 kg

Tfexpected = 6.55 -(20.20) * (11.2/128.1705) /0.35

Tfexpected = 1.5067

Tf mix = 1.28°C

relativ eerror --> (Real - Theoretical)/theoretical * 100 = (1.5067-1.28)/1.5067 * 100 = 15.04 %