Question

What is the freezing point of radiator fluid that is 50% antifreeze by mass? k f for water is 1.86 ∘ c/m?

Answer 1

Ethylene glycol is termed as the primary ingredients in antifreeze.
The ethylene glycol molecular formula is C₂H₆O₂.
Molar mass of C₂H₆O₂ is = (2×12) +(6×1) + (216) = 62g/mol
Now that antifreeze by mass is 50%, then there is 1kg of ethylene glycol which is present in 1kg of water.
ΔTf = Kf×m
ΔTf = depression in the freezing point.
= freezing point of water freezing point of the solution
= O°c - Tf
= -Tf
Kf = depression in freezing constant of water = 1.86°C/m
M is the molarity of the solution.
=(mass/molar mass) mass of solvent in kg
=1000g/62 (g/mol) /1kg
=16.13m
If we plug the value we get 
-Tf = 1.86 × 16.13 = 30
Tf = -30°c

Answer 2

Answer:

-30.02 ºC

Explanation:

Assuming the antifreeze to be ethylene glycol (C₂H₆O₂) which is popular antifreeze.

Molar mass of ethylene glycol (C₂H₆O₂) = 62g/mol

Step 1: calculate the freezing point depression of the solution

ΔT = -Kf*M

where,

ΔT= depression in the freezing point.

M = the molarity of the solution (mol solute / Kg solvent)

Kf = molar freezing point constant  of water = 1.86°C/m

To determine depression in the freezing point (ΔT), first we need to calculate;

• molarity of solute (ethylene glycol) in mol
• mass of solvent (water) in kg
• molarity of the solution (water +ethylene glycol)

Step 2: calculate the molarity of the solute (ethylene glycol)

Molar mass ethylene glycol = 62 g/mol

molarity of ethylene glycol in mol = 50 g / 62g/mol = 0.807 mol

Step 3: calculate mass of solvent in kg

There is 1kg of ethylene glycol which is present in 1kg of water

mass of solvent (water) in kg= 50 g/ 1000 g/ Kg = 0.050 Kg

Step 4: calculate the molarity of the solution (M)

M = 0.807 mol / 0.050 Kg = 16.14 m

Step 5: calculate the freezing point depression of the solution  (ΔT)

ΔT = - Kf*M = -1.86 ºC/m x 16.14 m

     = -30.02 ºC