Mixture

[Guest]

An amount of distilled water is added to 2 quarts of a 25% iodine solution to make the result a 20% solution. Determine amt. of water added.

25% of iodine = 25% of any quantity is 25% pure iodine. (as in this case 0.5 quarts).

Let x = amt. added.

x + 2 = amt. of quarts in mixture.

0.25 = 0.20 (2 + x)

0.25 = 0.4 + 0.20x

0.25 - 0.4 = 0.4 - 0.4 + 0.20x

-0.15 = 0.20x

-0.15 / 0.20 = 0.20x / 0.20

-0.75 = x

Where am I correct ?

[Guest]

Almost right.

The equation 0.25 = 0.20 (2+x) is wrong, all the other steps were correct though (assuming the equation was right).

amount of Iodine in the old solution = amount of iodine in the new solution (because only water is added)
amount of iodine = concentration [tex]\times[/tex] total amount
So
amount of iodine in the old solution = 0.25 [tex]\times[/tex] 2
amount of iodine in the new solution = 0.20 [tex]\times[/tex] (2+x)
So the equation is
0.25 [tex]\times[/tex] 2 = 0.20 (2+x)

Now when you do the rearranging of the equation (just like you did before) you should get
0.5 = x

Just to double check, this means:
the original solution is 0.5 quarts iodine out of 2 quarts, giving a concentration of 0.5/2 = 0.25.
the new solution is 0.5 quarts iodine out of 2.5 quarts, giving a concentration of 0.5/2.5 = 0.2.

Hope this helped,

R. Baber.

[Guest]

That helped. Thanks again.

[leesajohnson]

Yes, you are correct there is no mistake in it.