We are learning Colligative properties in class, I am fine with finding the freezing and boiling points etc. But when he gave us this question, I got really confused on how to approach it because it is asking for volume... Here it is. Thanks

You are planning on driving north to ski over Christmas break and you know that the coldest temperature may get is -10.0 degrees C. As you live in SC, your radiator is filled only with water, therefore, you want to add the correct amount of antifreeze to protect your car's radiator, which has a volume of 7.00L. Pure water freezes at exactly 0 degrees C and has a density= 1.000 g/ml. Water has a freezing temp of 1.86 degrees C/m. Antifreeze, MW 62.068 g/mol has a density of 1.1132 g/ml.

--What volume of antifreeze and what volume of water must be mixed together in order to have a total volume of solution that is 7.00L and which will not freeze until the temperature drops below -10.0 degrees C?

From Wikepedia

ΔT_{F}=K_{F}·b·i whereΔT_{F}= change in freezing point,K_{F}= freezing point constant,b= molality andi= van't Hoff factor (=1 for covalent substance like antifreeze)You have all but molality So 10 = 1.86 x b x 1 so molality = 10/1.86 = 5.376m, which means 5.376 moles antifreeze per kg of solvent or 5.376 x 62.068 g per kg of solvent = 333.68g antifreeze per kg water. 333.68g of antifreeze will have a volume of 333.68/1.1132 = 299.75mL of antifreeze per kg (1000g or 1000mL) of water. This makes a total volume of 1299.75 mL and this contains 297.75 mL of antifreeze. If 1299.75 mL of solution contains 297.75 mL of antifreeze, then 7L (7000mL) of this solution would contain 1603.6 mL of antifreeze (I think). Checkthis out and see if it makes any sense.

From memory of car 25% is often used and this would be a similar value (1750 ml).