9 1 pt The normal freezing point of benzene is 55 C The free

9. (1 pt) The normal freezing point of benzene is +5.5 °C. The freezing point depression constant, Kfpt, for this solvent is-5.12 °C/m. A solution is prepared by dissolving 55.0 g naphthalene (non-ionizing molecule, molar mass = 128. 17 g/mol) in 85.0 g benzene. What will the new freezing point be, for this solution? T, freexing point

Solution

Determine the mole(s) of naphthalene as

Mole(s) of naphthalene = (mass of naphthalene take)/(molar mass of naphthalene) = (55 g)/(128.17 g/mol) = 0.4291 mole.

We have 85.0 g benzene = (85.0 g)*(1 kg/1000 g) = 0.085 kg benzene.

Molality of the solution = (moles of naphthalene)/(kg of benzene) = (0.4291 mole)/(0.085 kg) = 5.0482 m.

Use Raoult’s law

T_f = K_f.pt*(molality of solution) where T_f = depression in freezing point of the solution and K_f.pt = -5.12°C/m. Plug in values and obtain

T_f = (-5.12°C/m)*(5.0482 m) = -25.8468°C.

We know that T_f = T_s – T_p

where T_s and T_p are the freezing points of the solution and pure benzene respectively. Put T_p = +5.5°C and obtain

-25.8468°C = T_s – (+5.5°C)

====> T_s = -25.8468°C + 5.5°C = -20.3468°C -20.35°C.

The freezing point of the solution is -20.35°C (ans).