Question 900095: How many liters of pure water should be mixed with an 8-L solution of 60% acid to produce a mixture that is 50% water? Found 2 solutions by stanbon, richwmiller:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! How many liters of pure water should be mixed with an 8-L solution of 60% acid to produce a mixture that is 50% water?
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Equation:
water + water = water
1*x + 0.40*8 = 0.50(x+8)
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100x + 40*8 = 50x + 50*8
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50x = 10*8
x = 8/5 = 1.6 litres (amt. of pure water to add)
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Cheers,
Stan H.
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You can put this solution on YOUR website! From the view of acid it is
since 100% water has 0% acid
0*x+.6*8=0.5(8+x)
4.8=4+.5x
.8=.5x
.8/.5=x
x=1.6 L
view of water
I changed from percent acid to percent non acid
60% acid means 40 % non acid
50% acid means 50 % non acid
Pure water is 100% non acid as a decimal 100%=1
1*x+0.4*8=0.5(8+x)
1*x+3.2=4+0.5x
1x-0.5x=4-3.2
0.5x=0.8
x=1.6 L Same answer!!!!
check
1*1.6+0.4*8=0.5(8+1.6)
1.6+3.2=0.5(9.6)
4.8=4.8
ok
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