Assuming 100% dissociation, calculate the freezing point and boiling point of 2.11 m Na2SO4(aq). Constants may be found here.
Solvent | Formula | Kf value*
(°C/m) |
Normal freezing point (°C) |
Kb value
(°C/m) |
Normal boiling point (°C) |
water | H2O | 1.86 | 0.00 | 0.512 | 100.00 |
benzene | C6H6 | 5.12 | 5.49 | 2.53 | 80.1 |
cyclohexane | C6H12 | 20.8 | 6.59 | 2.92 | 80.7 |
ethanol | C2H6O | 1.99 | –117.3 | 1.22 | 78.4 |
carbon tetrachloride |
CCl4 | 29.8 | –22.9 | 5.03 | 76.8 |
camphor | C10H16O | 37.8 | 176 |
Tf = i x Kf x m
Na2SO4 -----------> 2Na+ + SO4-2
giving 3 ions so i = 3
Kf = 1.86 0c/m
m = 2.11 m
Tf = 3 x 1.86 x 2.11
Tf = 11.8
as freezing point will decrease up on addition of solute
Tf = 0 - 11.8
Tf = -11.80C
Assuming 100% dissociation, calculate the freezing point and boiling point of 2.11 m Na2SO4(aq). Constants may...
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