how do you make 400 ml pf 2 Osm solution of Mg CL2?

m is molality and M is molarity; however, I understand Osm is the osmolarity of the solution. I will answer on that assumption.

MgCl2 has a van't Hoff factor of 3; i.e., MgCl2 ionizes in solution to three particles.
How many OSm do you want? That's 0.4 x 2/3 = ?
grams = ? x molar mass MgCl2.

To make a 400 mL 2 Osm solution of MgCl2, you first need to know the molar concentration of MgCl2 and its molecular weight.

The formula weight of MgCl2 is calculated as follows:
Mg (24.31 g/mol) + 2Cl (2 * 35.45 g/mol) = 95.21 g/mol

Next, we need to calculate the amount of MgCl2 needed to make the desired 2 Osm solution.

Osmolarity (Osm) is a unit of concentration representing the number of osmoles of solute per liter of solution. In this case, 2 Osm means 2 osmoles per liter. Since you want to make a 400 mL solution, you'll need to adjust the values accordingly.

First, calculate the number of moles of MgCl2 needed for the desired osmolarity:
Number of moles = Osmolarity * Volume (in liters)
Number of moles = 2 * (400 mL / 1000 mL/L)
Number of moles = 0.8 moles

Once you know the number of moles, you can calculate the mass of MgCl2 needed using the molecular weight:

Mass (in grams) = Number of moles * Molecular weight
Mass (in grams) = 0.8 moles * 95.21 g/mol
Mass (in grams) = 76.168 grams

Therefore, to make a 400 mL 2 Osm solution of MgCl2, you would need to measure 76.168 grams of MgCl2 and dissolve it in enough solvent (such as water) to make a total volume of 400 mL.