Question

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

Answer 1

To make a 500 mL of a 1.50 M BaCl2 solution from a 2.0 M barium chloride solution, you will need to measure out 375 mL of the 2.0 M solution and add enough water to bring the total volume up to 500 mL.

Step-by-step explanation:

To make a 500 mL of a 1.50 M BaCl2 solution from a 2.0 M barium chloride solution, you can use the dilution equation: M1V1 = M2V2. Rearranging the equation, we have V1 = (M2V2) / M1. Plugging in the given values, V1 = (1.50 M)(500 mL) / 2.0 M = 375 mL.

To make the dilution, you will need to measure out 375 mL of the 2.0 M barium chloride solution and add enough water to bring the total volume up to 500 mL. This will result in a 1.50 M BaCl2 solution.

Answer 2

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

Step-by-step 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.