How much ethylene glycol (C2H6O2, the major component of antifreeze) must be added to 1 L of water to keep it from freezing at -16 oC?Kf = 1.86 oC/m.
freezingptdepression=Kf*molality
-16=-1.86*molesglycol/1kg
solve for moles glycol
then convert that to grams.
To find out how much ethylene glycol must be added to 1 L of water to prevent freezing at -16°C, we need to calculate the concentration of ethylene glycol required.
First, let's determine the change in temperature (ΔT) from the freezing point of water (0°C) to the desired temperature (-16°C):
ΔT = -16°C - 0°C = -16°C
Next, we can use the formula for freezing point depression to calculate the molal concentration of the solute (ethylene glycol):
ΔT = Kf * m
Where:
ΔT = change in temperature in °C
Kf = cryoscopic constant for the solvent (water) = 1.86 °C/m
m = molal concentration of the solute (ethylene glycol) in mol/kg
Rearranging the equation, we can solve for the molal concentration:
m = ΔT / Kf
m = -16°C / 1.86 °C/m
m ≈ -8.60 mol/kg
Since we need to find the amount of ethylene glycol required in 1 L of water, we need to convert the molal concentration to mass:
1 kg of water = 1000 g
Therefore, the required mass of ethylene glycol can be calculated using its molar mass:
Molar mass of ethylene glycol (C2H6O2):
1 carbon atom (C) = 12.01 g/mol
6 hydrogen atoms (H) = 1.01 g/mol * 6 = 6.06 g/mol
2 oxygen atoms (O) = 16.00 g/mol * 2 = 32.00 g/mol
Molar mass of ethylene glycol = 12.01 g/mol + 6.06 g/mol + 32.00 g/mol = 62.07 g/mol
Next, we can use the equation:
mass = molar mass * molal concentration * mass of water
mass = 62.07 g/mol * -8.60 mol/kg * 1000 g
mass ≈ -533.14 g
Since having a negative mass makes no sense physically, we can disregard the negative sign and take the absolute value of the mass.
Therefore, approximately 533.14 grams of ethylene glycol should be added to 1 L of water to keep it from freezing at -16°C.