[CHM 112] Nickel Dilutions

[Andrew J. Pounds]

I have had this question come up a few times, so let me provide an example.

Let's say you weighed 3.9845 g of the nickel sulfate hexadrate. To find
the molarity of the first solution you find the moles of nickel sulfate
hexahydrate and then dilute it to 100 ml in the volumetric flask.

$M = \frac{\frac{3.9845}{262.9}}{0.100 L} = 0.1516\ M$

To make the second solution, you use 20 ml of this standard solution and
dilute it in a 50 ml volumetric flask.   In this case, use the
$M_1V_1=M_2V_2$ equation in a rearranged format.

$M_2 = \frac{M_1V_1}{V_2} = \frac{(0.1516 M)(0.02000 L)}{(0.05000 L)} =
0.06064 M$

You then use the same process, i.e. $M_1V_1=M_2V_2$, with varying
volumes to compute the molarities of solutions 3, 4, and 5.  Be sure and
use the correct molarity of the initial solution when you do the
calculation for the fifth solution.

-- 
Andrew J. Pounds, Ph.D.  (pounds_aj at mercer.edu)
Professor of Chemistry and Computer Science
Director of the Computational Science Program
Mercer University,  Macon, GA 31207   (478) 301-5627

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