Answer:
36 micrograms of [tex]Cu^{2+}[/tex] ion are therein one spot.
Explanation:
Amount of copper(II) ions in 1 liter solution = 6 g
Volume of solution used in spotting = 6μL = [tex]6\times 10^{-6} L[/tex]
[tex]1 \mu L = 10^{-6} L[/tex]
Amount of copper (II) ion in [tex]6\times 10^{-6} L[/tex]:
[tex]6\times 6\times 10^{-6} L=3.6\times 10^{-5} g[/tex]
[tex]1 g = 10^{6} \mu g[/tex]
[tex]3.6\times 10^{-5} g=3.6\times 10^{-5}\times 10^{6} \mu g=36 \mu g[/tex]
36 micrograms of [tex]Cu^{2+}[/tex] ion are therein one spot.
