Question

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?

Answer 1

`200" "mL` of 10% solution.
`300" "mL` 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="volume of " 10%" "(mL)`
`y="volume of " 20%" "(mL)`

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

`x+y=500" "mL`

`rArry=500-x` `" "(1)`

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:

`"mass in " x" " mL=C*V" "(g/(cancel(mL)))*cancel(mL)`

So in `500" mL"` we have `0.16*500" " (g/(cancel(mL)))*cancel(mL)=80" "g` of NaCl.

So, the sum of the masses of NaCl in `x` and `y` must equal `80" "g`, leading to the following expression:

`C_x*V_x+C_y*V_y=80`
`0.1x+0.2y=80` `" "(2)`

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

`0.1x+0.2(500-x)=80`

`0.1x-0.2x+100=80` `rArr0.1x=20` `rArrx=200" "mL`

Now solve for y using (1):

`y=500-x=500-200=300" "mL`

Answer 2

A different approach! Very detailed explanation given.

For the 20% constituent: `3/5xx500= 300" millilitres"`
For the 10% constituent: `2/5xx500=200" millilitres"`

Explanation:

`color(blue)("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.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("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

`1/(20-10)-=x/(16-10)" "` where `-=` means proportional to

`1/10=x/6`

Multiply both sides by 6

`6/10=x`

`x=3/5` of the whole
......................................................................................
Thus there is:

`3/5` of the 20% material

`2/5` of the 10% material

Check:

`[3/5xx20%] + [2/5xx10%] = 12+4 = 16%`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the volume of each constituent")`

For the 20% constituent: `3/5xx500= 300" millilitres"`
For the 10% constituent: `2/5xx500=200" millilitres"`