Analytical Chemistry
posted by Anonymous on .
25.0 mL of a 0.1067 M Solution of nickel (II) sulfate is diluted to 250 mL with water. 10 mL of this solution is taken and diluted to 1.0 L.
Calculate the concentration of [Ni2+] in the final solution in ppm.

0.1067M x (25/250) x (10/1000) = ?
or you can do this in you head.
25 to 250 is 1/10 and 10 to 1000 is 1/100 so the final concn is just 1/1000 x 0.1067 = 0.0001067M