Question

The boiling point of water is 100.00 °C at 1 atmosphere.

A student dissolves 12.70 grams of chromium(III) sulfate, Cr₂(SO₄)₃ (392.2 g/mol), in 214.5 grams of water. Use the table of boiling and freezing point constants to answer the questions below.

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

The molality of the solution is ___?____m.

The boiling point of the solution is ___?___°C.

Answer 1

1)

Lets calculate molality first

Molar mass of Cr2(SO4)3,

MM = 2*MM(Cr) + 3*MM(S) + 12*MM(O)

= 2*52.0 + 3*32.07 + 12*16.0

= 392.21 g/mol

mass(Cr2(SO4)3)= 12.70 g

use:

number of mol of Cr2(SO4)3,

n = mass of Cr2(SO4)3/molar mass of Cr2(SO4)3

=(12.7 g)/(3.922*10^2 g/mol)

= 3.238*10^-2 mol

m(solvent)= 214.5 g

= 0.2145 kg

use:

Molality,

m = number of mol / mass of solvent in Kg

=(3.238*10^-2 mol)/(0.2145 Kg)

= 0.151 molal

Answer: 0.151 molal

2)

i for Cr2(SO4)3 is 5 as it dissociates into 1 Cr3+ and 3 SO42-.

lets now calculate ΔTb

ΔTb = i*Kb*m

= 5.0*0.512*0.151

= 0.3865 oC

This is increase in boiling point

boiling point of pure liquid = 100.0 oC

So, new boiling point = 100 + 0.3865

= 100.3865 oC

Answer: 100.39 oC