SOLUTION: How many gallons of a 24% saline solution (NaCl) must be mixed with 5 gallons of pure water to make a 9% solution?

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Question 853076: How many gallons of a 24% saline solution (NaCl) must be mixed with 5 gallons of pure water to make a 9% solution?
Found 2 solutions by ewatrrr, josgarithmetic:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

Let x represent the 24% saline solution (76% water

 water = water

  .76x + 1.00(5gal) = .91(x + 5gal)

        .09(5gal)/.15  = x

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
The densities may be needed.
24%, about 1.19 g/ml
9%, about 1.07 g/ml
WATER, 1.00 g/ml and 1 gallon = 3.7854 liters

5 gallons of water is 5%2A3.7854=18.927 liters

The question could be restated as , how many liters of 24% w/vol saline must be mixed with 18.927 liters of water to make 9% w/vol. solution?
Start the solution process assigning variables:
u = volume liters of the 24% w/vol. saline

WRONG--SEE BELOW!___24%28grams%2F%28100%2Aml%29%29%281000%2Aml%2Fliter%29%2Au%2F%2818.927%2Bu%29=9 -------densities seem not to be needed.

Change the result for u back to gallons from the liters.
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The attempt at unit checking was overdone and created confusion. Since densities were not needed, the more typical setup should simply be:
highlight%28%2824u%29%2F%2818.927%2Bu%29=9%29
and this will give u=11.356 liters, which converts to 3 gallons.
RESULT: highlight%283%2Agallons%29 of 24% w/v saline.