Calculate the boiling point elevation of the given electrolytic solution:
5.7 g CaCl2 (calcium chloride) in 48.6 kg H2O(water)
Round your answer to 3 decimal places.
Solution :
Elevation in boiling point : It is a phenomenon that occurs when the Boiling point of a liquid (solvent ) is increased on the addition of a non-volatile compound.
It is a colligative property which means it depends on the 'number of solute' particles in the solution.
The formula for Elevation in Boiling point is
ΔTb = i * Kb * m -------------------- Equation 1
where ‘ i ’ is the van’t hoff factor of solute.
Kb = boiling point elevation constant of solvent
m = molality of the solution.
Here CaCl2 is the solute and water is the solvent.
Kb of water = 0.512 0C/m
Step 1 : To find the van’t hoff factor ‘i ‘
We are not given the van’t hoff factor in the question. So we will consider complete dissociation of CaCl2 .
CaCl2 ----> 1 Ca2+ (aq) + 2Cl- (aq)
Each unit of CaCl2 gives 1 ion of Ca2+ and 2 ions of Cl- ,
so van’t hoff factor ‘ i ’ = 1+2 = 3
Step 2 : To find molality of the solution
mass of CaCl2 (solute ) = 5.7g
mass of water (solvent ) = 48.6kg
molar mass of CaCl2 = 1 * atomic mass of Ca + 2 * atomic mass of Cl
= 1 * 40.078g/mol + 2 * 35.453g/mol
= 110.984g/mol
moles of CaCl2 = 5.7g CaCl2 * 1 mol CaCl2 / 110.984g CaCl2 = 0.05136 mol CaCl2
molality = moles of solute/ ‘kg’ of solvent = 0.05136mol CaCl2 /48.6kg
= 0.001057mol/kg or 0.001057m
Substitute i = 3, m = 0.001057m and Kb = 0.512 0C/m in the Boiling point Elevation formula (Equation 1 ), we have
ΔTb = i * Kb * m = 3 * 0.512 0C/m * 0.001057m
ΔTb = 0.001623 0 C.
In question we are told to write our answer to 3 decimal places.
So the boiling point elevation = 0.002 0C
Note : In the question we were not given van't factor 'i' . So we considered the complete dissociation of CaCl2 and consider i = 3. In case if you are given the van't hoff factor then instead of 3 you should use the given van't hoff factor.
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