You have to make 500 mL of a 1.50 M BaCl2. You have 2.0 M barium chloride solution available. Determine how to make the needed dilution

There are needed 375mL of the 2.0M BaCl₂ solution completing to 500mL with water.

Explanation:

We can find with the volume and concentration of the barium chloride the moles of BaCl₂ required. With the moles and the concentration of our stock solution we can know the volume of the 2.0M BaCl₂ solution required as follows:

Moles required:

0.500L * (1.50mol / L) = 0.750 moles BaCl₂

Volume stock solution:

0.750 moles BaCl₂ * (1L / 2.0mol) = 0.375L

h let me think

58750.08068 is the answer i think