To complete the question “how much water must be added to 100 cc of 80…” we need the usual full text:

How much water must be added to 100 cc of 80% solution of boric acid to reduce it to a 50% solution?

The answer is: 60 cc of water.

Quick Scoop

  • Initial solution: 100 cc of 80% boric acid.
  • Amount of boric acid present:
    • 80%80%80% of 100 cc = 80 cc of boric acid (this stays constant when you add water).
  • Let xxx = cc of water added.
    • Final volume = 100+x100+x100+x cc.
    • Final concentration required = 50%, so

80100+x=0.5\frac{80}{100+x}=0.5100+x80​=0.5

* Solving gives x=60x=60x=60.

So, 60 cc of water must be added to 100 cc of an 80% solution of boric acid to reduce it to a 50% solution.

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