Question

A 50.0mL Semple of benzene at 19.9°C was/ Cooled to its melting point, 5.500 and then frozen. Calculate how much heat transfer to the Starriounding occurred in this process, Density of benzene & 0.88oglmol, Chenzone 1745 g-6-1 Al pension - 9.920K5 MM = 78.114 g/mol) (9) How much energy is required to cool the benzene from 19.9°C to 5.5°C b. blhat is the enthalpy of freezing? Calculate the amount et energy required to freeze this amount of benzene. What is the total amount of energy required? -

Answer 1

(a) The energy required to cool benzene = {50 mL * 0.88 g/mL * 1.74 J/g.^oC * (19.9-5.5) ^oC} + {(50 mL * 0.88 g.mL^-1/78.114 g.mol^-1)*9920 J/mol) = 6.69 kJ

(b) Enthalpy of freezing = 6.69 kJ/(50 mL * 0.88 g.mL^-1/78.114 g.mol^-1) = 11.877 kJ/mol