In Duluth, Mn the average temperature is -13.10 deg C . How much ethonal glycol, C2H6O2, must be dissolved in 1.00 kg of water to make a radiator fluid that wouldn't freeze until the temp reached -20.00 deg C?
would my answer be 666.5 g C2H6O2 ?
To calculate the amount of ethylene glycol (C2H6O2) required to prevent the water from freezing at a specific temperature, you will need to use the concept of freezing point depression and the freezing point depression constant of water.
Freezing point depression occurs when a solute (in this case, ethylene glycol) is added to a solvent (water), causing the freezing point of the mixture to decrease. The relationship between the freezing point depression and the molality of the solute is given by the equation:
ΔT = Kf * m
Where:
ΔT is the change in freezing point (in this case, -20.00°C - (-13.10°C) = -6.90°C),
Kf is the freezing point depression constant of water (for water, Kf = 1.86°C/m),
m is the molality of the ethylene glycol solution.
To find the molality (m), we need to convert the given mass of ethylene glycol to moles and the mass of water to kilograms.
1. Calculate the moles of ethylene glycol:
Given mass of ethylene glycol = 666.5 g
Molar mass of ethylene glycol (C2H6O2) = 46.07 g/mol
moles of ethylene glycol = mass / molar mass
moles of ethylene glycol = 666.5 g / 46.07 g/mol = 14.45 mol
2. Calculate the mass of water in kilograms:
Given mass of water = 1.00 kg
3. Calculate the molality:
molality (m) = moles of solute / mass of solvent (in kg)
molality (m) = 14.45 mol / 1.00 kg = 14.45 m
4. Use the freezing point depression equation to find ΔT:
ΔT = Kf * m
ΔT = 1.86°C/m * 14.45 m = 26.91°C
Therefore, the freezing point of the water is lowered by 26.91°C when 666.5 g of ethylene glycol is dissolved in 1.00 kg of water.
So, your answer of 666.5 g of C2H6O2 is correct.