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To what volume should you dilute $\pu{126 mL}$ of an $\pu{8.15 M}$ $\ce{CuCl2}$ solution so that $\pu{49.5 mL}$ of the diluted solution contains $\pu{4.56 g}$ $\ce{CuCl2}?$

I know you need to use $c_1V_1=c_2V_2,$ but do not know how to incorporate the $\pu{4.56 g}$ of $\ce{CuCl2}$ into it.

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    $\begingroup$ First figure out the concentration of the resultant, 49.5 mL contains 4.56 g CuCl2, then starting with 8.15 molar solution, how much must it be diluted. $\endgroup$ Feb 9 at 0:41

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I figured I need to use resulting molar concentration:

$$c_2 = \frac{m_i}{M(\ce{CuCl2})\cdot V_i} = \frac{\pu{4.56 g}}{(\pu{134.45 g mol-1})(\pu{0.0495 L})} = \pu{0.685 M}$$

$$V_2 = \frac{c_1V_1}{c_2} = \frac{(\pu{8.15 M})(\pu{126 mL})}{\pu{0.685 M}} \approx \pu{1.50 L}$$

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    $\begingroup$ I took a liberty to brush up formatting, notations and significant figures. Keep in mind it's advantageous to first solve the problem using symbolic algebra, and plug in the values at the very last step. Also, please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Finally, refrain from asking in the answering section — if you post an answer, it shouldn't be a proofreading request. $\endgroup$
    – andselisk
    Feb 9 at 5:46

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