Question 516357: a chemist has a 10 ml of a 30% acid solution. how many millilters of pure acid must be added to obtain a 50% solution? Answer by Maths68(1474) (Show Source):
You can put this solution on YOUR website! Soluton A
Amount = 10 ml
Concentration = 30%=0.3
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Solution B
Amount = x ml
Concentration = 100%=1 (Pure acid)
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Resultant Solution
Amount=10+x ml
Concentration=50%=0.5
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(Amount of A * Concentration of A)+(Amount of B * Concentration of B) = (Amount of Resultant * Concentration of Resultant)
(10*0.3)+(x*1)=(10+x)*0.50
3+x=5+0.5x
x-0.5x=5-3
0.5x=2
0.5x/0.5=2/0.5
x=4 ml
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10 ml of a 30% acid solution and 4 ml of pure acid must be mixed to obtain 14 ml of 50% solution
