Question

1. Show quantitatively how you would prepare a solution of 3.10 x 10^2 g of 0.125 molar (m) ethylene alcohol from ethylene glycol and water.

Answer 1

A CAUTIOUS note here. m stands for molality. M stands for molarity. If 0.125 is molal (m) the problem is easy enough; if 0.125 M (molar) you need the density which is not given. In the absence of definitive instructions, I will assume you intend to prepare a 0.125 molal solution.
1. First, I have no idea what ethylene alcohol is. However, here is what you do.

2. m = moles/kg solvent
0.125 = moles/0.310
Solve for moles (of solute).
3. moles solute = grams solute/molar mass solute. Solve for grams solute.