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 110 mg of pentane, C5H12, in 16.45 g of cyclohexane. Assume an ideally dilute solution and that only pure cyclohexane freezes out.

Answer 1

Molar mass of cyclohexane = 84.16 g/mole

Molar mass of cyclopentane = 70.1 g/mole

GIVEN:

Freezing point of the pure cyclohexane solution, T_f⁰ = 6.47 ⁰C = 6.47+273.15 = 279.62 K

Mass of cyclohexane = 16.45 g

Mass of cyclopentane = 110 mg = 0.11 g

Heat of fusion of cyclohexane, ∆H_fus = 31.3 J/g = (31.3 J/g) (84.16 g/mole)

                                                                  = 2634.208 J/mole

No. of moles of cyclohexane, n₁ = 16.45g / (84.16 g/mole) = 0.1955799 moles

No. of moles of cyclopentane, n₂ = 0.11g/ (70.1 g/mole) = 0.001569 moles

Now, ∆T = R(T_f⁰)² X_solute

                              ∆H_fus

                                     OR

m=molality , M_A = molar mass of cyclohexane

Here, ∆T = T_f⁰ – T_f

          R = 8.3144 J mol^-1K^-1

          X_solute = mole fraction of solute i.e cyclopentane =     n₂ / (    n₁ + n₂) .          

                                                                              = 0.007958

So, ∆T = (8.3144 J mol^-1K^-1) (279.62 K)²(0.007958)

                         (2634.208 J/mole)

     T_f⁰ – T_f = 1.9639 K

     279.62 – T_f = 1.9639 K

      T_f = 279.62 – 1.9639 = 277.656 K

       T_f = 4.506 ⁰C

Freezing point of the solution of 110 mg of pentane in 16.45 g of cyclohexane = 4.5 ⁰C