Question 17
Actual freezing point of water = 0 °C
The following formula can be used
Δt = i Kf m
Kf for water is 1.86 °C/m
i = 3 for MgCl2 it produces three ions in the water
m = molality
Moles of MgCl2 = 10.74 g / 95.211 g/mol = 0.1128 moles
Molality = 0.1128 mol / 0.08659 Kg = 1.3027
Δt = 3 x 1.86 °C/m x 1.3027 mol / Kg
Δt = 7.269 °C
Frrezing point = 0 °C- 7.269 °C= -7.269 °C
The solution freezes -9.26 °C
Question 18
Actual freezing point of water = 0 °C
The following formula can be used
Δt = i Kf m
Kf for water is 1.86 °C/m
i = 2 for NaCl it produces two ions in the water
m = molality
Moles of MgCl2 = 16.31 g / 58.45 g/mol = 0.279moles
Molality = 0.279 mol / 0.09318 Kg = 2.9946 mol/kg
Δt = 2 x 1.86 °C/m x 2.9946 mol / Kg
Δt =11.14 °C
Frrezing point = 0 °C- 11.14 °C= -11.14 °C
The solution freezes -11.14 °C
Question 19
Actual freezing point of water = 0 °C
The following formula can be used
Δt = i Kf m
Kf for water is 1.86 °C/m
i = 1 for nonelectrolysis including sugar
m = molality
Moles of Sugar = 46.18 g / 342.3 g/mol = 0.1349 moles
Molality = 0.1349 mol / 0.09093 Kg = 1.483 mol/kg
Δt = 1 x 1.86 °C/m x 1.483 mol / Kg
Δt =2.759 °C
Frrezing point = 0 °C- 2.759 °C= -2.759°C
The solution freezes -2.759 °C
17 Marks: 1 What is the freezing point of water made by dissolving 10.74 g of...
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