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?

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  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

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