Question cba82

May 31, 2016

You need $317.8 \text{ml}$ of the 15% solution and $332.2 \text{ml}$ of the 60% solution.

Explanation:

You need to use the idea that the concentration of a solute = amount of solute / volume of solution.

In symbols:

$c = \frac{n}{v} \text{ } \textcolor{red}{\left(1\right)}$

The total amount of solute before and after the dilution remains the same.

Rearranging $\textcolor{red}{\left(1\right)}$ gives:

$n = c \times v \text{ } \textcolor{red}{\left(2\right)}$

Now we can set up 2 simultaneous equations from the information given. We have 2 equations and 2 unknowns so we can solve it.

From $\textcolor{red}{\left(2\right)}$ we can write:

$\left(60 \times {V}_{60}\right) + \left(15 \times {V}_{15}\right) = 650 \times 38$

$\therefore 60 {V}_{60} + {V}_{15} = 24 , 700 \text{ } \textcolor{red}{\left(3\right)}$

Where ${V}_{60}$ and ${V}_{15}$ refer to the volumes of the 60% and 15% solutions respectively.

I am using the %# as the concentration.

Since we are told the total volume we can write:

${V}_{60} + {V}_{15} = 650 \text{ } \textcolor{red}{\left(4\right)}$

So $\textcolor{red}{\left(3\right)}$ and $\textcolor{red}{\left(4\right)}$ are our simultaneous equations.

Rearranging $\textcolor{red}{\left(4\right)} \Rightarrow$

${V}_{60} = 650 - {V}_{15}$

We can now substitute this value for ${V}_{60}$ back into $\textcolor{red}{\left(3\right)} \Rightarrow$

$60 \left(650 - {V}_{15}\right) + 15 {V}_{15} = 24 , 700$

$\therefore 39 , 000 - 60 {V}_{15} + 15 {V}_{15} = 24 , 700$

$\therefore 60 {V}_{15} - 15 {V}_{15} = 39 , 000 - 24 , 7000$

$\therefore 45 {V}_{15} = 14 , 300$

$\therefore {V}_{15} = \frac{14 , 300}{45} = \textcolor{red}{317.8 \text{ml}}$

Substituting this value back into $\textcolor{red}{\left(4\right)} \Rightarrow$

${V}_{60} + 317.8 = 650$

$\therefore {V}_{60} = 650 - 317.8 = \textcolor{red}{332.2 \text{ml}}$