The boiling point of water is
100.00 °C at 1 atmosphere.
A student dissolves 13.51 grams of sodium
sulfate, Na2SO4
(142.1 g/mol), in 262.4 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
mass(solute)= 13.51 g
use:
number of mol of solute,
n = mass of solute/molar mass of solute
=(13.51 g)/(1.421*10^2 g/mol)
= 9.507*10^-2 mol
m(solvent)= 262.4 g
= 0.2624 kg
use:
Molality,
m = number of mol / mass of solvent in Kg
=(9.507*10^-2 mol)/(0.2624 Kg)
= 0.3623 molal
Answer: 0.3623 molal
2)
i for Na2SO4 is 3 as 1 molecule of Na2SO4 dissociates into 2 Na+ and 1 SO42- ions
lets now calculate ΔTb
ΔTb = i*Kb*m
= 3.0*0.512*0.3623
= 0.5565 oC
This is increase in boiling point
boiling point of pure liquid = 100.0 oC
So, new boiling point = 100 + 0.5565
= 100.56 oC
Answer: 100.56 oC
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