Question

For cyclohexane, C₆H₁₂, the normal melting point is 6.47°C and the heat of fusion at this temperature is 31.3 J/g. Find the freezing point of a solution of 188 mg of pentane, C₅H₁₂, in 16.45 g of cyclohexane. Assume an ideally dilute solution and that only pure cyclohexane freezes out.

R= 8.3145 J/mol.K, T(K)=T(°C)+273.15.

The molecular weight of pentane is 72.15 g/mol

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

Tb mix = Tb solvent - dTb

then

mol of pentane = mass/MW = 0.188/72.1488 = 0.00261

kg of solvent = 16.45*10^-3 kg

molal = mol/kg = 0.00261 /( 16.45*10^-3 ) =  0.15866

now,

Kf = -20.2 °C/m

dTf = -KF*m = -20.2*0.15866 = -3.2049

Tf = 6.47 + -3.2049

Tf = 3.2651 °C