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

`m_(H_2O)=312.7g\\m_(CaCl_2)=164.703g`

Step-by-step explanation:

Hello,

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

`\%m/m=(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.

Answer 2

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

Step-by-step explanation:

Let c = mass of calcium chloride

and w = mass of water

You have two conditions:

`\begin{array}{rcl}\mathbf{(1)}\quad 34.5 & = & (c )/(c + w) * 100\\\\\mathbf{(2)}\ c+ w & = & 477.4\\w & = & 477.4 - c\\\\34.5 & = & (c )/(c + 477.4 - c) * 100\\\\34.5 & = & (c )/(477.4)* 100\\\\c & = & (34.5* 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}}`