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<div class="moz-cite-prefix">On 6/14/19 7:27 PM, wrote:<br>
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<div>For the nickel quantification lab, to calculated the
concentration(M) for solutions 2-5, wouldn’t we need to first
know the amount of solution 1 within each and calculate it
that way. I am a little confused. Could you please lead me in
the right direction for these calculations, thanks!</div>
<div><br>
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</blockquote>
<p><br>
</p>
<p>I have had this question come up a few times, so let me provide
an example.</p>
<p>Let's say you weighed 3.9845 g of the nickel sulfate hexadrate.</p>
<p>To find the molarity of the first solution you find the moles of
nickel sulfate hexahydrate and then diluted it to 100 ml in the
volumetric flask.</p>
<p><img alt="$M = \frac{\frac{3.9845}{262.9}}{0.100 L} = 0.1516\ M$"
style="vertical-align: middle;"
src="cid:part1.0FA4C962.2CCBBE8A@mercer.edu"><br>
</p>
<p>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 <img alt="$M_1V_1=M_2V_2$" style="vertical-align:
middle;" src="cid:part2.F4906DDD.5E705174@mercer.edu"> equation
in a rearranged format.</p>
<p><img alt="$M_2 = \frac{M_1V_1}{V_2} = \frac{(0.1516 M)(0.02000
L)}{(0.05000 L)} = 0.06064 M$" style="vertical-align: middle;"
src="cid:part3.35684569.98759DA9@mercer.edu"></p>
<p>You then use the same process, i.e. <img alt="$M_1V_1=M_2V_2$"
style="vertical-align: middle;"
src="cid:part2.F4906DDD.5E705174@mercer.edu">$, 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.</p>
<p><br>
</p>
<pre class="moz-signature" cols="72">--
Andrew J. Pounds, Ph.D. (<a class="moz-txt-link-abbreviated" href="mailto:pounds_aj@mercer.edu">pounds_aj@mercer.edu</a>)
Professor of Chemistry and Computer Science
Director of the Computational Science Program
Mercer University, Macon, GA 31207 (478) 301-5627
</pre>
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