Mixture (or solution) problems use equations to relate the amounts (either mass or volume) of each substance used in the mixture. The total amount in a mixture can be expressed as:
where \(x\) and \(y\) represent the amount of each individual substance.
There can be many different substances in a mixture. For example, if you were to mix \(4\) fruits into a fruit salad, the total amount would be expressed as \(A = w + x + y + z\).
The total amount of the slime mixture is the sum of the different substances:
\( A = w + b + g \)
\( A = \cfrac{1}{4} + \cfrac{1}{4} + \cfrac{1}{8} \)
In order to add these substances more easily, we can give their respective fractions a common denominator:
\(A = \cfrac{2}{8} + \cfrac{2}{8} + \cfrac{1}{8} \)
\( A = \cfrac{5}{8} \; [\text{cups}]\)
Therefore, we can determine that this recipe makes \(\boldsymbol{\cfrac{5}{8} \; [\textbf{cups}]}\) of slime.
Mixtures can have quantities in weight or volume. Some examples are shown below:
| \(\text{Weight}\) | \(\text{g, kg, lb, ton, oz}\) |
| \(\text{Volume}\) | \(\text{mL, L, m³, cups}\) |
When the substances added into the mixture are mixtures themselves, we may be interested in only a certain substance within the mixture. The amount of a specific substance in a mixture is \(a = w \cdot A\). Where \(a\) is the amount of the substance, \(w\) is the weight/volume percentage of the substance in the mixture and \(A\) is the amount of the mixture.
Rubbing alcohol can contain up to \(70\%\) pure alcohol. This means, up to \(70\%\) of the volume (or mass) is pure alcohol (\( w \)). Since we know the volume of the full bottle (\( A \)), we can use this information to calculate the volume of pure alcohol (\( a \)):
\( a = w \cdot A \)
\( a = 70\% \cdot 250 \)
\( a = 0.70 \cdot 250 \)
\( a = 175 \; [\text{mL}] \)
The total amount of alcohol in the running alcohol mixture is \(\textbf{175 [mL]}\).
The question asks for the volume of \(30\%\) and \(90\%\) hydrochloric acid that can be mixed together to make \(100\;[\text{mL}]\) of \(36\%\) hydrochloric. This information is summarized in the table below.
| Solution | \(A\) | \(w\) |
|---|---|---|
| \(30% \text{HCl}\) | \(x\) | \(30\%\) |
| \(90% \text{HCl}\) | \(y\) | \(90\%\) |
| \(36% \text{HCl}\) | \(100\) | \(36\%\) |
Here, \(x\) and \(y\) represent the volume in \([\text{mL}]\) of the \(30\%\) and \(90\%\) \(\text{HCl}\) respectively.
We know that the the total amount \(A\) in the \(36\%\) solution is \(100 \; [\text{mL}]\) which is made up of \(x\) and \(y\). We also know that the amount of HCl in a given solution is \(a = w \cdot A\).
The total HCl in the \(36\%\) solution is \(36 \% \cdot 100\). This is made up from the HCl in solutions \(x\) and \(y\) which is \(30 \% \cdot x\) and \(90 \% \cdot y\).
| \(\text{Amount}\) | \(x + y = 100\) |
| \(\text{HCl}\) | \((30 \% \cdot x) + (90 \% \cdot y) = (36 \% \cdot 100)\) |
First, we can rearrange the first equation to isolate for \(x\):
\(x = 100 - y\)
Next, we can simplify the second equation:
\((30 \% \cdot x) + (90 \% \cdot y) = (36 \% \cdot 100)\)
\((0.30 \cdot x) + (0.90 \cdot y) = (36)\)
Then, we can substitute the first equation into second equation:
\((0.30(100 - y)) + (0.90y) = 36\)
\((30 - 0.30y) + (0.90y) = 36\)
\(- 0.30y + 0.90y = 36 - 30\)
\( 0.60y = 6\)
\( y = \cfrac{6}{0.60}\)
\( y = 10 \; [\text{mL}]\)
Finally, we can plug \(y\) into the first equation to determine \(x\):
\(x = 100 - y\)
\(x = 100 - 10\)
\(x = 90 \; [\text{mL}]\)
Therefore, to make a \(100\;[\text{mL}]\) solution of \(36\%\) HCl, we need to mix \(\textcolor{red}{\textbf{90 [mL]}}\) of \(\textcolor{red}{\textbf{30% HCl}}\) and \(\textcolor{blue}{\textbf{10 [mL]}}\) of \(\textcolor{blue}{\textbf{90% HCl}}\). It seems reasonable that the \(35\%\) solution is made up mostly of \(30\%\) solution with a some \(90\%\) solution.
Try another example here.