Question

what volume (in ml) of water would have to be added to 150.0ml of 0.450 m potassium chloride, kcl, to give a solution with a concentration of 0.100 m?

Answer 1

Solution.

$n(KCl) = 0.15 \times 0.450 = 0.0675 \ mol$

$V'(KCl) = \frac{n(KCl)}{C'(KCl)}$

V'(KCl) = 0.675 L = 675 ml

$V(H2O) = 675 - 150 = 525 ml$

Answer:

V(H2O) = 525 ml