# Suitable antifreeze solution

If glycerine $\ce{C3H5(OH)3}$and methyl alcohol $\ce{CH3OH}$ are sold at the same price per kilogram, which would be cheaper for producing an antifreeze solution for the radiator of an automobile radiator?

I think that more depression in freezing point will result in a better antifreeze solution. It depends on molal freezing constant. How can I predict this?

• I think the key point of this (probably homework) question, is about the units of price per kilogram, while a product is sold per liter. The density of glycerol is 1.261 g/cm3 and of methanol 0.792 g/cm3 (Wikipedia). So with equal price per kilogram, the price per liter of glycerol is 59% higher than of methanol Oct 7, 2016 at 9:18
• You should use the molar mass of the compounds instead of their density Oct 7, 2016 at 9:37
• @Lucademics the price is per kilogram and not per litre
– Pink
Oct 7, 2016 at 10:12
• Since modern automobile engines contain aluminum, and methanol corrodes aluminum, it should not be used as antifreeze at any price! You might also check if even glycerol requires inhibitors to prevent corrosion. Oct 9, 2016 at 2:51

You are on the right track when you mention molality.

Because freezing point depression is a colligative property of the solvent, in this case water, all you need to do is to find out which solute has more moles per kg, aka molality.

The equation for freezing point depression for a non-ionic solute is given as:

$$\Delta T_\mathrm{F} = K_\mathrm{F}\times b$$

Where:
$\Delta T_\mathrm{F} =$ freezing point depression
$K_\mathrm{F} =$ cryoscopic constant (dependent on solvent, not solute)
$b =$ molality, the moles of solute per kg of solvent

The key point here is that freezing point depression is linearly proportional to solute concentration (molality).

Since glycerol has a molecular mass of $\pu{91.1 g/mol}$, 1 kg will contain 11.0 moles of solute. Methanol has a molecular mass of $\pu{32.0 g/mol}$, so 1 kg will contain 31.2 moles of solute.

So, for any given price per kg, methanol will give you $31.2/11.0 = 2.9$ times as many moles of solute, resulting in a 2.9 times greater freezing point depression. Note that you cannot calculate the actual freezing point depression without knowing $K_\mathrm{F}$.