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ser-zykov
13 days ago
7

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

Physics
2 answers:
ValentinkaMS [3.3K]13 days ago
7 0

Response:

-30.02 ºC

Explanation:

Assuming the antifreeze used is ethylene glycol (C₂H₆O₂), which is a commonly used antifreeze.

The molar mass of ethylene glycol (C₂H₆O₂) is calculated to be 62g/mol

Step 1: determine the freezing point depression for the solution

ΔT = -Kf*M

where,

ΔT= the decrease in freezing point.

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

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

To find the decrease in freezing point (ΔT), we first need to compute;

  • molarity of the solute (ethylene glycol) in moles
  • mass of solvent (water) measured in kg
  • molarity of the solution (water mixed with ethylene glycol)

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

Molar weight for ethylene glycol = 62 g/mol

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

Step 3:calculate the mass of the solvent in kg

1kg of ethylene glycol is mixed with 1kg of water

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

Step 4:calculate the solution's molarity (M)

M = 0.807 mol / 0.050 Kg = 16.14 m

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

ΔT = - Kf*M = -1.86 ºC/m x 16.14 m

     = -30.02 ºC

Maru [3.2K]13 days ago
7 0
Ethylene glycol is known as the main component found in antifreeze.
The molecular formula for ethylene glycol is C₂H₆O₂.
Its molar mass is calculated as C₂H₆O₂ = (2×12) +(6×1) + (216) = 62g/mol
Given that antifreeze comprises 50% by weight, there exists 1 kg of ethylene glycol mixed with 1 kg of water.
ΔTf = Kf×m
ΔTf refers to the change in the freezing point.
= starting temperature of water - freezing temperature of the solution
= 0°C - Tf
= -Tf
Kf stands for the freezing point depression constant of water, which is 1.86°C/m
m is the molarity of the solution.
=(mass/molar mass) where mass of solvent is in kg
=1000g/62 (g/mol) /1kg
=16.13m
Substituting the value into the equation gives us
-Tf = 1.86 × 16.13 = 30
thus Tf = -30°C
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