SOLUTION:
What quantity of 70 per cent acid solution must be mixed with a 15 percent solution to produce 4 mL of a 30 per cent solution?
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What quantity of 70 per cent acid solution must be mixed with a 15 percent solution to produce 4 mL of a 30 per cent solution?
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Question 693487:
What quantity of 70 per cent acid solution must be mixed with a 15 percent solution to produce 4 mL of a 30 per cent solution? Found 2 solutions by ankor@dixie-net.com, checkley79:Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! What quantity of 70 per cent acid solution must be mixed with a 15 percent solution to produce 4 mL of a 30 per cent solution?
:
Let x = quantity of 70% solution required
then resulting total amt is 4, therefore:
(4-x) = quantity of 15% solution
:
.70x + .15(4-x) = .30(4)
.70x + .6 - .15x = 1.2
.70x - .15x = 1.2 - .6
.55x = .6
x = .6/.55
x ~ 1.1 liters of 70% solution required
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:
:
Check this
.7(1.1) + .15(4-1.1) = .3(4)
.77 + .435 = 1.2
You can put this solution on YOUR website! .70X+.15(4-X)=.30*4
.70X+.60-.15X=1.20
.55X=1.20-.60
.55X=.60
X=.60/.55
X=1.09 ML OF THE 70% SOLUTION IS USED.
4-1.09=2.91 ML OF THE 15% SOLUTION IS USED.
PROOF:
.70*1.09+.15*2.91=1.20
.763+.436=1.20
1.1995=1.20
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