Question

The freezing point of water is 0.00°C at 1 atmosphere.  

A student dissolves 12.62 grams of aluminum acetate, Al(CH₃COO)₃ (204.1 g/mol), in 254.5 grams of water. Use the table of boiling and freezing point constants to answer the questions.

1. What is the molality of the solution?

2. What is the freezing point of the solution?

Solvent	Formula	Kb (°C/m)	Kf (°C/m)
Water	H2O	0.512	1.86
Ethanol	CH3CH2OH	1.22	1.99
Chloroform	CHCl3	3.67
Benzene	C6H6	2.53	5.12
Diethyl ether	CH3CH2OCH2CH3	2.02

Answer 1

1.

molality    = weight of solute*1000/gram molecular weight of solute* weight of solvent in g

                = 12.62*1000/(204.1*254.5)

                = 0.243m

Al(CH₃COO)₃ (aq) ----------> Al^3+ (aq) + 3CH3COO^- (aq)

i   = 4

Tf   = i*Kf*m

          = 4*1.86*0.243

          = 1.81⁰C

Tf    = Tf solvent - Tf solution

1.81 = 0 -Tf solution

Tf solution = -1.81⁰C

The freezing point of the solution    = -1.81⁰C