# A chemist mixes a 10% saline solution with a 20% saline solution to make 500 milliliters of a 16% saline solution. How many milliliters of each solution does the chemist mix together?

Jun 13, 2017

$200 \text{ } m L$ of 10% solution.
$300 \text{ } m L$ of 20% solution.

#### Explanation:

To solve for two unknowns we need two equations/pieces of information. Our two unknowns are the volumes of each stock solution. Let:

$x = \text{volume of " 10%" } \left(m L\right)$
$y = \text{volume of " 20%" } \left(m L\right)$

For our first equation, we know the total volume is 500 mL and is the sum of x and y:

$x + y = 500 \text{ } m L$

$\Rightarrow y = 500 - x$$\text{ } \left(1\right)$

For our second equation, we do a mass balance for 500 mL of final solution.

16%" w/v"=0.16" "g/(mL)

This means that in 1 mL of solution, we have 0.16 g of NaCl.

For any solution, concentration multiplied by volume will give the mass of NaCl:

$\text{mass in " x" " mL=C*V" } \left(\frac{g}{\cancel{m L}}\right) \cdot \cancel{m L}$

So in $500 \text{ mL}$ we have $0.16 \cdot 500 \text{ " (g/(cancel(mL)))*cancel(mL)=80" } g$ of NaCl.

So, the sum of the masses of NaCl in $x$ and $y$ must equal $80 \text{ } g$, leading to the following expression:

${C}_{x} \cdot {V}_{x} + {C}_{y} \cdot {V}_{y} = 80$
$0.1 x + 0.2 y = 80$$\text{ } \left(2\right)$

Now, substitute our expression for x, (1), into (2):

$0.1 x + 0.2 \left(500 - x\right) = 80$

$0.1 x - 0.2 x + 100 = 80$$\Rightarrow 0.1 x = 20$$\Rightarrow x = 200 \text{ } m L$

Now solve for y using (1):

$y = 500 - x = 500 - 200 = 300 \text{ } m L$

Jun 13, 2017

A different approach! Very detailed explanation given.

For the 20% constituent: $\frac{3}{5} \times 500 = 300 \text{ millilitres}$
For the 10% constituent: $\frac{2}{5} \times 500 = 200 \text{ millilitres}$

#### Explanation:

$\textcolor{b l u e}{\text{Preamble about method}}$

The total volume is a fixed amount. So if the proportion of the 20% concentration is known then the amount of 10% solution is:

" "color(brown)("volume of 10% = total fixed volume - volume of 20%")

Thus by just focusing on the 20% the amount of the 10% is indirectly linked. Thus in this approach we can (sort of) forget about the amount of 10% mix.

By varying the amount of the 20% mix the saline content of the whole changes. It is this change that we are looking at.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine the proportions of each constituent}}$

Let the proportion of the 20% solution be $x$

The gradient of part is the same as the gradient of the whole.

Using ratio:" " ("proportion of 20% concentration")/("saline content in blend") ->1/(20-10)

Giving

$\frac{1}{20 - 10} \equiv \frac{x}{16 - 10} \text{ }$ where $\equiv$ means proportional to

$\frac{1}{10} = \frac{x}{6}$

Multiply both sides by 6

$\frac{6}{10} = x$

$x = \frac{3}{5}$ of the whole
......................................................................................
Thus there is:

$\frac{3}{5}$ of the 20% material

$\frac{2}{5}$ of the 10% material

Check:

[3/5xx20%] + [2/5xx10%] = 12+4 = 16%
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine the volume of each constituent}}$

For the 20% constituent: $\frac{3}{5} \times 500 = 300 \text{ millilitres}$
For the 10% constituent: $\frac{2}{5} \times 500 = 200 \text{ millilitres}$