How many gallons of ethylene glycol, molar mass 62.07 g/mol, density = 1.1 g/mol do you need to add to each gallon of water ( 1 gallon = 4 liters and kf = 1.85 C/m) in your radiator to get the water to freeze at -15 C?
Is it 0.458 gallon?
Looks ok to me.
To calculate the amount of ethylene glycol needed to add to each gallon of water, we can follow these steps:
1. Determine the molality of the solution:
Molality (m) = moles of solute / mass of solvent (in kg)
Since we want to know how much ethylene glycol to add to each gallon of water, let's assume we have 1 gallon of water. From the given information, we know that the density of water is 1.1 g/mL, so 1 gallon of water is equivalent to 4 liters or 4000 mL.
Mass of water = volume × density = 4000 mL × 1.1 g/mL = 4400 g
We need to use 4400 g of water because the given quantity is 1 gallon (which is approximately 4 liters).
2. Convert the mass of water to moles:
Moles of water = mass / molar mass = 4400 g / 18.015 g/mol = 244.53 mol
3. Calculate the freezing point depression (ΔTf):
ΔTf = kf × m
The given kf value is 1.85 °C/m and we want the water to freeze at -15 °C. The freezing point of pure water is 0 °C. Therefore, the ΔTf = 0 °C - (-15 °C) = 15 °C.
Rearranging the formula, we can solve for m:
m = ΔTf / kf = 15 °C / 1.85 °C/m = 8.11 m
4. Compute the moles of ethylene glycol needed:
Moles of ethylene glycol = m × moles of water = 8.11 m × 244.53 mol = 1986.08 mol
5. Convert moles of ethylene glycol to mass:
Mass of ethylene glycol = moles × molar mass = 1986.08 mol × 62.07 g/mol = 123,275.05 g
6. Convert mass of ethylene glycol to gallons:
Since 1 gallon is equivalent to 3.78541 L, we can convert the mass to gallons:
Gallons of ethylene glycol = mass (g) / density (g/mL) / volume of 1 gallon (L) × 1000 mL/L / 1000 g/g = 123,275.05 g / 1.1 g/mL / 3.78541 L/gallon × 1000 mL/L / 1000 g/g = 30.25 gallons (approximately)
Therefore, the correct answer is approximately 30.25 gallons of ethylene glycol needs to be added to each gallon of water to achieve a freeze point of -15 °C.
So, the answer provided (0.458 gallon) is incorrect.