Word Problems - Mixtures 1

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 is \(A = x + y\). Where x and y are the amount of each individual substance. There can be many substance in a mixture. Supose you mix 4 fruits into a fruit salad. The total amount will be \(A = w + x + y + z\).

To make slime, add \( \frac{1}{4} \) cup of water, \( \frac{1}{4} \) cup of borax and \( \frac{1}{8} \) cup of glue. How many cups of slime does this recipe make?

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Mixtures can have quantities in weight or volume. Some examples are shown below:

\(\text{Weight}\)	\(\text{g, kg, lb, ton, oz}\)
\(\text{Volume}\)	\(\text{mL, L, m^3, cups}\)

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 \(70%\) of pure isopropyl alcohol. In a \(250\;[mL]\) bottle, how much pure alcohol is in it?

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A chemical reaction calls for \(100\;[mL]\) of \(36%\) hydrochloric acid but the only solutions available are \(30%\) and \(90%\) hydrochloric acid. How much of each are required to make the solution you need?

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The question asks for the volume of \(30\%\) and \(90\%\) hydrochloric acid that can be mixed together to make \(100\;[mL]\) 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 \([mL]\) of the \(30\%\) and \(90\%\) \(\text{HCl}\).

We know that the the total amount \(A\) in the \(36\%\) solution is \(100 \; [mL]\) 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 \(90 \% \cdot y\).

\(\text{Amount}\)	\(x + y = 100\)
\(\text{HCl}\)	\((30 \% \cdot x) + (90 \% \cdot y) = (36 \% \cdot 100)\)

Let's rearrange the first equation to isolate for \(x\):

\(x = 100 - y\)

Next, substitue into equation 2:

\((30 \% \cdot x) + (90 \% \cdot y) = (36 \% \cdot 100)\)

\((0.30 \cdot x) + (0.90 \cdot y) = (36)\)

\((0.30 \cdot (100 - y)) + (0.90 \cdot y) = 36\)

\((30 - 0.30 \cdot y) + (0.90 \cdot y) = 36\)

\( - 0.30 \cdot y + 0.90 \cdot y = 36 - 30\)

\( 0.60 \cdot y = 6\)

\( y = \frac{6}{0.60}\)

\( y = 10 \; [mL]\)

Now plug in \(y\) back into the first equation:

\(x = 100 - y\)

\(x = 100 - 10\)

\(x = 90 \; [mL]\)

Therefore, to make a \(100\;[mL]\) solution of \(36\%\) HCl, we need to mix \(90\;[mL]\) of \(30\%\) HCl and \(10 \;[mL]\) of \(90\%\) HCl. It seems reasonable that the \(35\%\) solution is made up mostly of \(30\%\) solution with a some \(90\%\) solution.

Try another example here.

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