Achemist makes 240. ml of zinc nitrate (zn(no) working solution by adding distilled water to 70.0 ml of a 2.50stock solution of zinc nitrate in water. calculate the concentration of the chemist's working solution. round your answer to 3 significant digits.

M = 0.729 M

Explanation:

In this case, we have a solution of Zn(NO₃)₂.

The chemist wants to prepare a dillute solution of this reactant.

The stock solution of the nitrate has a concentration of 2.5 M, and he wants to prepare 240 mL of a more dillute concentration of the same solution. He adds 70 mL of the stock and complete with water until it reach 240 mL.

We want to know the concentration of this dilluted solution.

As we are working with the same solution, we can assume that the moles of the stock solution will be conserved in the dilluted solution so:

n1 = n2  (1)

and we also know that:

n = M*V (2)

If we replace this expression in (1) we have:

M1*V1 = M2*V2

Where 1, would be the stock solution and 2, the solution we want to prepare.

So, we already know concentration and volume used of the stock solution and the desired volume of the dilluted one, therefore, all we have to do is replace the given data in (2) and solve for the concentration which is M2:

2.5 * 70 = 240M2

M2 = 2.5 * 70 / 240

This is the concentration of the solution prepared.

the answer is c i took the test