Question

A solution is made by dissolving 0.585 mol of nonelectrolyte solute in 877 g of benzene. Calculate the freezing point, Tf, and boiling point, Tb, of the solution. Constants can be found in the table of colligative constants.

Constants for freezing-point depression and boiling-point elevation calculations at 1 atm:

Solvent	Formula	Kf value* (°C/m)	Normal freezing point (°C)	Kb value (°C/m)	Normal boiling point (°C)
water	H2O	1.86	0.00	0.512	100.00
benzene	C6H6	5.12	5.49	2.53	80.1
cyclohexane	C6H12	20.8	6.59	2.92	80.7
ethanol	C2H6O	1.99	–117.3	1.22	78.4
carbon tetrachloride	CCl4	29.8	–22.9	5.03	76.8
camphor	C10H16O	37.8	176

*When using positive K_f values, assume that ΔT_f is the absolute value of the change in temperature. If you would prefer to define ΔT_f as "final minus initial" temperature, then ΔT_f will be negative and so you must use negative K_f values. Either way, the freezing point of the solution should be lower than that of the pure solvent.

Tf= ∘C

Tb=   ∘C

Answer 1

T_f = 2.075°C

T_{​​​​​​b} = 81.79°C