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