I need help to dilute insecticide. How much water to add??

Question

I have a 27.5 oz bottle of insecticide and the active chemical ingredient is at 21.4%.    I need to dilute the active chemical ingredient down to 9%.  How much water should I add to 27.5 oz to make the active ingredient 9% ? 

Thank you.  Edn

 

Answer 1

 

A 27.5 oz. bottle of insecticide contains an active chemical at 21.4%

Let x = the amount (oz.) of water we must add to the bottle to dilute the active chemical in the bottle to 9%

Thus, the amount of active ingredient in the bottle = .214x27.5 oz. = 5.884 oz. of active ingredient in the bottle of insecticide.

 

If we want the reduce the % of active ingredient in the bottle, we must add an amount of water to the bottle as follows:

 

Goal: 5.884 oz. of the active ingredient in the bottle = 9% of the insecticide in the bottle.

 

Solution: 5.884 oz. of active ingredient in the bottle = 9% of (27.5 oz. + x oz.)

So:         5.884 = .09(27.5+x)

And:       5.884 = 2.475+.09x

So:         5.884-2.475 = 2.475-2.475+.09x

And:       3.409 = .09x

And:       3.409 =  .09x

                .09      .09  

And:         37.9 =   x

 

Therefore: the amount of water we must add to the insecticide = 37.9 oz.

Giving us a new total amount of insecticide in the bottle of 27.5 oz.+37.9 oz. = 65.4 oz.  

 

Check:

27.5 oz. of insecticide in the bottle + 37.9 oz. of water added to the bottle = 65.5 oz. insecticide in the bottle

Now the % of active ingredient in the bottle is

 

5.884 x 100 = .0899% 

65.4

which rounds to 9% active ingredient in the bottle