Question 248129
how many liters of water must be added to 8 liters of a 40% acid solution to obtain a 10% acid solution?
:
Let x = amt of water
a simple "amt of acid" equation (the amt of acid does not change only the percent)
.40(8) = .10(x+8)
3.2 = .10x + .8
3.2 - .8 = .1x
2.4 = .1x
x = {{{2.4/.1}}}
x = 24 liters of pure water required
:
Check for equal acid: 
.4(8) = .10(24+8)
3.2 = .1(32)
;
;
How many kg of pure salt must be added to 20 kilograms of a 10% salt solution to obtain a 25% salt solution?
:
Let x = amt of pure salt
.10(20) + x = .25(x+20)
2 + x = .25x + 5
x - .25x = 5 - 2
.75x = 3
x = {{{3/.75}}}
x = 4 kg of salt to be added
;
:
Check
.10(20) + 4 = .25(4+20)
2 + 4 = .25(24)
6 = 6