Question

How much CuCl2 is needed to prepare 325g of 1.00%(w/w) solution?

Answer 1

Answer on Question #49815 – Chemistry - Inorganic Chemistry

How much $\mathrm{CuCl}_2$ is needed to prepare $325\,\mathrm{g}$ of $1.00\%$ (w/w) solution?

Solution:

$m_{\mathrm{sol}} = 325\,\mathrm{g}$

$w(\mathrm{CuCl}_2) = 1.0\%$

$m(\mathrm{CuCl}_2) - ?$

$$w(\mathrm{CuCl}_2) = \frac{m(\mathrm{CuCl}_2)}{m_{\mathrm{sol}}} \times 100\%$$ $$m(\mathrm{CuCl}_2) = \frac{w(\mathrm{CuCl}_2) \times m_{\mathrm{sol}}}{100\%}$$ $$m(\mathrm{CuCl}_2) = \frac{1.00\% \times 325\,\mathrm{g}}{100\%} = 3.25\,\mathrm{g}$$

Answer: $3.25\,\mathrm{g}$ of $\mathrm{CuCl}_2$ is needed.

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