Question

How many microliters of a 904.0 ppm Co2 stock solution are required to prepare 19.00 mL of a 24.0 ng/μL Co2 solution?

Answer 1

Ans. Desired [CO₂] = 24.0 ng/ uL                                       ; [1 ng = 10^-6 mg ; 1 uL = 10^-6 L]

                                    = (24.0 x 10^-6 mg) / (10^-6 L)

                                    = 24.0 mg/ L                                     ; [1 ppm = 1 mg/ L]

                                    = 24.0 ppm

# So far we have-

            Desired solution = 24.0 ng/ uL = 24.0 ppm

            Stock solution = 904.0 ppm

Now, using-

            C1V1 (stock solution) = (C2V2) desired solution

            Or, 904.0 ppm x V1 = 24.0 ppm x 19.00 mL

            Or, V1 = (24.0 ppm x 19.00 mL) / 904.0 ppm = 0.50442 mL

            Hence, V1 = 0.50422 mL

So, required volume of stock solution = 0.50422 mL                            ; [1 mL = 1000 uL]

                                                            = 504.22 uL