You need to prepare 1.0 L of solution containing 1.0 mM MgCl2 and 0.15 M NaCl from stock solutions of 1.0 M MgCl2 and 0.50 M NaCl. You will:

mix 1 mL of the stock solution of MgCl2, 300 mL of the stock solution of NaCl and 699 mL of water.

Explanation:

We need to determine the volume necessary of both stock solutions. When a dilution is made, a certain volume of the stock solution is collected and then more solvent is added to it until the volume is completed. So, the number of moles of the solute in both solutions are the same, and it is the concentration (C) multiplied by the volume (V). If 1 is the stock solution and 2 the diluted solution:

C1*V1 = C2*V2

So, in this case, the two solutions will be mixed, and then the volume will be completed with the solvent. So, for MgCl2:

C1 = 1.0 M

C2 = 1.0 mM = 0.001 M

V2 = 1.0 L = 1000 mL

1*V1 = 0.001*1000

V1 = 1 mL of the stock solution of MgCl2

For NaCl:

C1 = 0.50 M

C2 = 0.15 M

V2 = 1000 mL

0.50*V1 = 0.15*1000

V1 = 300 mL

So, it will be necessary 1 mL of the stock solution of MgCl2, 300 mL of the stock solution of NaCl and 699 mL of water.