SOLUTION: You have 30 cl of acid solution that is 10% in strenght, how much water would you need to dilute the solution to 7% in strenght. I have fought with this task for hours now and I

10% of 30 is 3. So

30 cl of 10% solution contains 3 cl of pure acid [and the rest (27 cl) is
water.  Forget the water, just think about the amt of acid and the amt of
liquid].

So you have 30 cl of liquid that contains 3 cl of acid

Now you're going to add x liters of liquid (water).

So now you have 30+x cl of liquid that still contains only 3 cl of acid.

Now those 3 cl of acid must be that 7% of 30+x.
          |                 |       |  |
So        3                 =     .07  *  (30+x)


                         3 = .07*(30+x)

Multiply both sides by 100 to get rid of the decimal

                       300 = 7*(30+x)
                       300 = 210+7x
                        90 = 7x
                         = x

That works out to about 12.9 cl of water to dilute it down to 7%

Checking.  You'll end up with about 30+12.9 or 42.9 cl. and 7% of
that is 42.9(.07) = 3.003, and that's close enough to 3, and we 
expected it be a little off because the 12.9 was rounded off.


Edwin