# Finding final volume of the diluted solution with defined mass concentration given initial molar concentration

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.

• 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. Feb 9 at 0:41

$$c_2 = \frac{m_i}{M(\ce{CuCl2})\cdot V_i} = \frac{\pu{4.56 g}}{(\pu{134.45 g mol-1})(\pu{0.0495 L})} = \pu{0.685 M}$$
$$V_2 = \frac{c_1V_1}{c_2} = \frac{(\pu{8.15 M})(\pu{126 mL})}{\pu{0.685 M}} \approx \pu{1.50 L}$$