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.
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
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