ok, i'm given:

Ethylene Glycol C2H6O2 is the main ingredient in antifreeze.
density of pure EG: 1.132 g/mL
Anti-freeze is 32% by volume EG

I have to find:
Volume of Eg in 1 L
Mass of Eg in 1 L
& the Volume of "solvent" in 1 L

Can someone help me? and give me some tips? For some reason, this certain topic is KILLING me.

Of course, I can help you with that! Let's break down the information step by step and find the answers to your questions.

First, let's find the volume of Ethylene Glycol (EG) in 1 liter of antifreeze.

We know that antifreeze is 32% by volume EG. So, in 1 liter (1000 mL) of antifreeze, 32% of it will be EG. Let's calculate it:

Volume of EG in 1 L = 32% of 1000 mL
Volume of EG in 1 L = (32/100) * 1000 mL
Volume of EG in 1 L = 320 mL

Therefore, the volume of EG in 1 liter of antifreeze is 320 mL.

Next, let's find the mass of EG in 1 liter of antifreeze.

We are given that the density of pure EG is 1.132 g/mL. Since we now know that the volume of EG in 1 L of antifreeze is 320 mL, we can calculate the mass.

Mass of EG in 1 L = density * volume
Mass of EG in 1 L = 1.132 g/mL * 320 mL
Mass of EG in 1 L = 363.04 g

So, the mass of EG in 1 liter of antifreeze is 363.04 grams.

Now, let's find the volume of the solvent (the other liquid component besides EG) in 1 liter of antifreeze.

To find this, we can subtract the volume of EG from the total volume of antifreeze.

Volume of solvent in 1 L = Total volume - Volume of EG
Volume of solvent in 1 L = 1000 mL - 320 mL
Volume of solvent in 1 L = 680 mL

Therefore, the volume of the solvent in 1 liter of antifreeze is 680 mL.

To summarize and answer your questions:
- The volume of EG in 1 liter of antifreeze is 320 mL.
- The mass of EG in 1 liter of antifreeze is 363.04 grams.
- The volume of the solvent in 1 liter of antifreeze is 680 mL.

I hope this explanation helps you understand how to approach this problem. If you have any further questions, feel free to ask!