When octane is used, the solution will have less effect on the freezing point depression of the solution
Explanation:
The complete question is:
Calculate the freezing point of a solution of 125 g KBr in 450 g water.
Since a solution of KBr would probably cause significant damage to the car radiator and engine, you decide to use 125 g of the nonelectrolyte octane (molar mass 114 g/mole). Will this have a greater or less effect on freezing point depression of the solution?
Step 1: Data given
Molar mass of KBr = 119.0 g/mol
Molar mass of octane = 114 g/mol
Mass of KBr = 125 grams
Mass of octane = 125 grams
Mass of water = 450 grams
Step 2: Calculate moles KBr
Moles KBr = mass KBr / molar mass KBr
Moles KBr = 125 grams / 119.0 g/mol
Moles KBr = 1.05 moles
Step 3: Calculate moles octane
Moles octane = 125 grams / 114 g/mol
Moles octane = 1.10 moles
Step 4: Calculate molality
Molality = moles compound / mass water
Molality KBr = 1.05 moles / 0.450 kg
Molality KBr = 2.33 molal
Molality octane = 1.10 moles / 0.450 kg
Molality octane = 2.44 Â molal
Step 5: Calculate the freezing point depression when KBr is used
ΔT = i*Kf * m
⇒with ΔT = the freezing point depression = TO BE DETERMINED
⇒with i = the van't Hoff factor of KBr = 2
⇒with Kf = the freezing point depression constant of water = 1.86 °C/m
⇒with m= the molality = 2.33 molal
ΔT = 2*1.86 * 2.33
ΔT = 8.68 °C
This means the freezing point of this solution is -8.68 °C
Step 6: Calculate the freezing point depression when octane is used
ΔT = i*Kf * m
⇒with ΔT = the freezing point depression = TO BE DETERMINED
⇒with i = the van't Hoff factor of the nonelectrolyte octane = 1
⇒with Kf = the freezing point depression constant of water = 1.86 °C/m
⇒with m= the molality = 2.44 molal
ΔT = 1* 1.86 * 2.44
ΔT = 4.54 °C
This means the freezing point of this solutions is -4.54 °C
When octane is used, the solution will have less effect on the freezing point depression of the solution
