# How much water must be added to 12 grams of a 90% iodine solution to produce a 25% iodine solution?

Aug 24, 2015

#### Answer:

You would have to add 31.2 g of water.

#### Explanation:

Iodine dissolves in water only to 0.03 % concentration, so this is a purely hypothetical question.

90 % iodine means $\text{90 g iodine"/"100 g solution}$

So in 12 g of 90 % solution:

$\text{Mass of iodine" = 12 color(red)(cancel(color(black)("g soln"))) × "90 g iodine"/(100 color(red)(cancel(color(black)("g soln")))) = "10.8 g iodine}$

We add $x \text{ g}$ of water to $\text{12 g}$ of 90 % iodine to get $12 + x \text{ g}$ of 25 % iodine.

$\frac{10.8}{12 + x} = \frac{25}{100}$

10.8×100=25(12+x)

$1080 = 300 + 25 x$

$25 x = 1080 - 300 = 780$

$x = \frac{780}{25}$

$x = 31.2$

We have to add 31.2 g of water.

Check:

$\frac{10.8}{12 + 31.2} = \frac{25}{100}$

$\frac{10.8}{43.2} = \frac{25}{100}$

$\frac{1}{4} = \frac{1}{4}$

It checks!