Question

The boiling point of an aqueous solution is 101.88 °C. What is the freezing point? Constants can be found here.

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

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

get dTb

dTb = 101.88-100 = 1.88°C

then

dTb = i*m*Kb

1.88 = i*m * 0.512

im = 1.88/0.512 = 3.671875

now, substitute in dTf and Kf

dTf = -Kf*molality * i

dTf = -Kf*3.671875

for water, 1.86

dTf = -1.86*3.671875

dTf = -6.829°C