Question

An experiment requires 250 mℓ of a 80 μM solution of potassium chloride. A stock solution

of concentration 45 mM is available. What volume of stock solution should be used? Your

answer should be expressed in μℓ to at least 3 significant figures.

Answer 1

Proportion:

$80 \times 10^{-6} mol$ – 1000 mL

x mol – 250 mL

$x = \frac{80 \times 10^{-6} \times 250}{1000} = 20 \times 10^{-6} \;mol$ (we need)

Proportion:

$45 \times 10^{-3}$ mol – 1000 mL

$20 \times 10^{-6}$ mol – y mL

$y = \frac{20 \times 10^{-6} \times 1000}{45 \times 10^{-3}} = 0.444 \;mL$

Answer: 444 μℓ