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 KgStep 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
