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.
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
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