Question

17 Marks: 1 What is the freezing point of water made by dissolving 10.74 g of magnesium chloride in 86.59 g of water? The freezing-point depression constant of water is 1.86 °CIm Answer: 18 Marks: 1 What is the freezing point of water made by dissolving 16.31 g of sodium chloride in 93.18 g of water? The freezing-point depression constant of water is 1.86 °C/m. Answer 19 Marks: 1 What is the freezing point of water made by dissolving 46.18 g of sugar (C12H22011) in 90.93 g of water? The freezing-point depression constant of water is 1.86 °C/m. Answer:

Answer 1

Question 17

Actual freezing point of water =  0 °C

The following formula can be used

Δt = i K_f m

K_f 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 K_f m

K_f 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 K_f m

K_f 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