Question

How many microliters of a 872.0 ppm Co2 stock solution are required to prepare 19.00 mL of a 13.0 ng/μL Co2 solution?

W/ Correct Sig Figs

Answer 1

According to law of dilution   MV = M'V'

Where M = Molarity of stock = 872.0 ppm = 872.0 mg/L

                                                             = 872.0x10^-3 g/10⁶ μL

                                                             = 872.0x10^-9 g/μL

                                                             = 872.0 ng/μL

V = Volume of the stock = ?

M' = Molarity of dilute solution = 13.0 ng/μL

V' = Volume of the dilute solution = 19.00 mL

Plug the values we get , V = ( M'V') / M

                                       = (13.0x19.0) / 872.0

                                      = 0.28 mL

Therefore the volume of stock solution required is 0.28 mL