1 liter of a mixture containing 60% of boric acid.How much water must b added 2 make de mixture 48% boric acid

Question

can u plesh show de work instead of just putting de answer? thx :]

Answer 1

Let x be the required amount of water to be added.

Balance by the concentration = amount of solute / total mixture

amount of solute: 0.6 liters
total mixture:0.4+x liters.

48% = 0.6/(1+x)

Solve for x,
x = 0.25 liters.

Answer 2

Take
(% x original amount) + (& x desired amount) = (& x total amount)
But make sure the % is of the same thing..
60% boric acid = 40% water
48% boric acid = 52% water
water = 100% water

.4(1) + 1(x) = .52(1 + x)

Now solve.

.4 + x = .52 + .52x
1x - .52x = .52 - .4
.48x = .12
x = .25

.25 liters must be added