Question

In order to perform a chemical reaction, 225 mL of 0.500 M lead (II) nitrate is required. How much 5.00 M stock solution is required to make the proper solution?

Answer 1

15.0 mL

Step-by-step explanation:

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Answer 2

15.0 mL x x. = 0.850 g KCl. 1000 mL mol KCl.

Step-by-step explanation:

EXAMPLE: What mass of potassium chloride would be needed to prepare 250.0 mL of a 0.500. M solution? 1 L.