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

neutrons are in the middle of the atom and the electrons, which are much smaller.