A radiator is to be protected against freezing to -20f. Calculate the volume of C2H4(OH)2 ethylene glycol per liter of water. Density of glycol is 1.11 g/ml
Is that -20 F? Convert to C. I get approximately -29 but you need to confirm that more accurately.
delta T = Kf*m
-29 = 1.86*m
Solve for m = about 15.
m = 15 mol/kg solvent
1 L has a mass of 1 kg; therefore, we need 15 mol ethylene glycol.
g ethylene glycol = mols x molar mass = 15 x about 62 g/mol = about 950 g ethylene glycol. Convert that to mL using the density of ethylene glycol.
To calculate the volume of ethylene glycol (C2H4(OH)2) per liter of water needed to protect a radiator against freezing to -20°F, we'll need to determine the concentration of glycol solution required to achieve this level of freeze protection.
First, let's convert -20°F to Celsius since most freeze protection charts reference temperatures in Celsius.
To convert Fahrenheit to Celsius, we can use the formula: °C = (°F - 32) × 5/9
So, for -20°F:
°C = (-20 - 32) × 5/9
= (-52) × 5/9
= -28.89°C
Now, referring to freeze protection charts for ethylene glycol, we can determine the concentration required to protect against freezing at -28.89°C. Let's assume we want to achieve a freeze protection level of around -20°F or -28.89°C.
Typically, freeze protection charts for ethylene glycol solutions provide information on the percentage of glycol required for different temperature ranges. For example, they may indicate that a 50% concentration by volume provides freeze protection down to -35°C (-31°F).
To calculate the volume of ethylene glycol per liter of water, we'll need to know the specific concentration required. You can refer to a glycol concentration/temperature chart or consult the manufacturer's recommendations for the desired freeze protection level.
Once you have the required glycol percentage, the calculation is straightforward.
For example, if a 50% concentration is needed, it means that half the volume is glycol (C2H4(OH)2) and the other half is water.
Thus, the volume of glycol per liter of water would be 0.5 liters.
If you know the density of the glycol, which is given as 1.11 g/ml, you can also determine the mass of glycol needed.
To calculate the mass of glycol per liter of water, we can multiply the density (1.11 g/ml) by the volume of glycol (0.5 L):
Mass of glycol = density × volume
= 1.11 g/ml × 0.5 L
= 0.555 g
Therefore, for a 50% ethylene glycol concentration, the volume of C2H4(OH)2 per liter of water is 0.5 liters, or the mass of C2H4(OH)2 per liter of water is 0.555 grams.