Question

A solution of CaCl2 in water forms a mixture that is 34.5% calcium chloride by mass. If the total mass of the mixture is 477.4 g, what masses of CaCl2 and water were used?

Answer 1

Answer:

$\boxed{\text{164.7 g of CaCl _{2 } and 312.7 g of water}}$

Explanation:

 Let c = mass of calcium chloride

and w = mass of water

You have two conditions:

$\begin{array}{rcl}\mathbf{(1)}\quad 34.5 & = & \dfrac{c }{c + w} \times 100\\\\\mathbf{(2)}\ c+ w & = & 477.4\\w & = & 477.4 - c\\\\34.5 & = & \dfrac{c }{c + 477.4 - c} \times 100\\\\34.5 & = & \dfrac{c }{477.4}\times 100\\\\c & = & \dfrac{34.5\times 477.4}{100}\\\\c & = &\mathbf{164.7}\\\\164.7 + w & = & 477.4\\w & = &\mathbf{312.7}\\\end{array}\\\text{The solution consisted of }\boxed{\textbf{164.7 g of CaCl _{2 } and 312.7 g of water}}$

Answer 2

Answer:

$m_{H_2O}=312.7g\\m_{CaCl_2}=164.703g$

Explanation:

Hello,

In this case, we take into account the by mass percentage as:

$\%m/m=\frac{m_{CaCl_2}}{m_{mixture}}$

In such a way, we compute the mass of calcium chloride which is contained into 477.4g of 34.5% mixture as:

$m_{CaCl_2}=m_{mixture}*\%m/m\\m_{CaCl_2}=0.345*477.4g=164.703g$

Finally, the mass of water is computed via the total mass minus the mass of calcium chloride:

$m_{H_2O}=m_{mixture}-m_{CaCl_2}\\m_{H_2O}=477.4g-164.703g\\m_{H_2O}=312.7g$

Best regards.