Help with the calculation

Question

I have 300 mL of a 1.5 M KCl solution. If I heat the solution until the volume of the solution is 250 mL,

Describe how you will prepare a 0.1 M 100.0 mL potassium chloride solution from solid potassium chloride.?

Answer 1

Not sure what the question is in the first part of your question. You boil away 50 mls from the original 300 ml of 1.5 M KCl. Are you asking what the new concentration is of the resulting 250 mls? If so, then here is one way to calculate that answer:

(300 ml)(1.5 M) = (250 ml)(x M)

x = 1.8 M KCl

You want to prepare 100.0 ml of 0.1 M KCl solution from solid KCl. Sine 0.1 M means 0.1 mol KCl per liter of solution, and we want only 100 mls (0.1 L), this means we want -- 0.1 L x 0.1 mol/L = 0.01 moles KCl

Molar mass KCl = 74.6 g / mol

0.01 moles KCl = 0.01 moles x 74.6 g / mole = 0.746 moles KCl needed

Since 0.1 M has only 1 signifiant figure, our answer should not have more than 1 s.f. Thus, we should report our answer as 0.7 g KCl needed