SOLUTION: a chemist needs 10 liters of a 25 % acid solution. the solution is to be mixed from three solutions whose acid concentrations are 10% 20% 50% how many liters of each solution shou
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: a chemist needs 10 liters of a 25 % acid solution. the solution is to be mixed from three solutions whose acid concentrations are 10% 20% 50% how many liters of each solution should the chemist use to satisfy the following
a use as little as possible the 50% solution
b use as much as possible of the 50% solution
c use 2 liters of the 50% solution this is a three part question This question is from textbook
You can put this solution on YOUR website! a chemist needs 10 liters of a 25 % acid solution. the solution is to be mixed from three solutions whose acid concentrations are 10% 20% 50% how many liters of each solution should the chemist use to satisfy the following
:
a use as little as possible the 50% solution
Mix it with 20% solution only
Let x = amt of 50% solution
then
(10-x) = 20% solution
equation:
.50(x) + .20(10-x) = .25(10)
.5x + 2 - .2x = 2.5
.3x = 2.5-2
x =
x = 1.67 liters of 50% solution required
10 - 1.67 = 8.33 liters of 20% solution
:
;
b use as much as possible of the 50% solution
Mix it with the 10% solution only
Let x = amt of 50% solution
then
(10-x) = 10% solution
equation:
.50x + .10(10-x) = .25(10)
.5x + 1 - .1x = 2.5
.4x = 2.5 - 1
x =
x = 3.75 liters of 50% solution
10 - 3.75 = 6.25 liters of 10% solution
;
:
c use 2 liters of the 50% solution
Mix it with 10% and the 20%; they will have to equal 8 liters
Let x = 20% solution
then
(8-x) = 10% solution
equation:
.20x + .10(8-x) + .50(2) = .25(10)
.20x + .8 - .10x + 1 = 2.5
.2x - .1x + 1.8 = 2.5
.1x = 2.5 - 1.8
.1x = .7
x =
x = 7 liters of 20% solution
(8-7 = 1 liter of the 10% solution
:
:
Check the last one:
.2(7) + .1(1) + .5(2) = .25(10)
1.4 + .1 + 1 = 2.5
