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.

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.

To prepare a solution with a desired molarity, you need to understand the relationship between molarity (M), moles (mol), and volume (L) of the solute.

In this case, you want to prepare a 0.125M solution of ethylene alcohol (C2H6O2) using ethylene glycol and water.

Step 1: Calculate the moles of ethylene alcohol needed.
To find the moles, you can use the equation:
moles = molarity × volume (in liters).

Since you want a specific mass of ethylene alcohol, you need to convert 3.10 × 10^2 g to moles. To do this:
moles of ethylene alcohol = mass / molar mass.

The molar mass of ethylene alcohol (C2H6O2) can be calculated by adding the atomic masses of each element in it:
C: 2 atoms × 12.01 g/mol = 24.02 g/mol
H: 6 atoms × 1.01 g/mol = 6.06 g/mol
O: 2 atoms × 16.00 g/mol = 32.00 g/mol

Therefore, the molar mass of C2H6O2 = 24.02 g/mol + 6.06 g/mol + 32.00 g/mol = 62.08 g/mol.

Now, you can calculate the moles of ethylene alcohol:
moles = 3.10 × 10^2 g / 62.08 g/mol.

Step 2: Calculate the volume of ethylene glycol needed.
Since ethylene alcohol will be prepared from ethylene glycol, you need to calculate the volume of ethylene glycol required.

The equation relating moles, molarity, and volume remains the same:
moles = molarity × volume (in liters).

In this case, you need to solve for volume, so rearranging the equation gives:
volume (in liters) = moles / molarity.

Substituting the values:
volume (in liters) = (3.10 × 10^2 g / 62.08 g/mol) / 0.125 mol/L.

Step 3: Calculate the volume of water needed.
To prepare the desired solution, you must take into account the final volume and subtract the volume of ethylene glycol.

Let's assume you want to prepare a final volume of 1 liter. Now apply the equation:
volume of water (in liters) = total volume - volume of ethylene glycol.

Substituting the values:
volume of water (in liters) = 1 L - volume of ethylene glycol.

Step 4: Prepare the solution.
Measure the calculated volumes of ethylene glycol and water using appropriate laboratory glassware, such as a measuring cylinder or volumetric flask.

Pour the calculated volume of ethylene glycol into a container, followed by the calculated volume of water. Mix them thoroughly to ensure a homogeneous solution.

By following these quantitative calculations and subsequent steps, you can prepare a solution of 3.10 × 10^2 g of 0.125 molar ethylene alcohol using ethylene glycol and water.