Question

Start with a 1040.0 ppm Stock Zn solution. Prepare control 1 by diluting 2.00 mL of Stock Zn solution in a 200.0 mL volumetric flask and dilute to the mark. Prepare control 2 by diluting 2.00 mL of control 1 into a 100.0 mL volumetric flask. What is the Zn concentration (ppm Zn) in control 2?

Answer 1

According to law of dilution   MV = M'V'

Where M = Molarity of stock = 1040.0 ppm

V = Volume of the stock = 2.0 mL

M' = Molarity of control 1 solution = ?

V' = Volume of the control 1 solution = 200.0 mL

Plug the values we get   , M' = MV /V'

= 10.4 ppm

For the control 2 solution ,

According to law of dilution   M"V" = M'V'

Where M = Molarity of control 1 = 10.4 ppm

V = Volume of the stock = 2 mL

M' = Molarity of dilute solution = ?

V' = Volume of the dilute solution = 100.0 mL

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

= 0.208 ppm

Therefore the concentration of Zn in control 2 solution is 0.208 ppm