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?

2 Answers
Jun 13, 2017

Answer:

#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#

Jun 13, 2017

Answer:

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")#

Tony B

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"#