Mixture: Amount to Be Added

[Guest]

A physician has a 10% solution of iodine that weighs 10 oz. He needs a 5% solution. How much alcohol must be added ?

Amount of ounces to be added: N (in partial solution to problem)

Amount of ounces in new solution: N + 10 (I added this)

Amount of ounces of iodine in first solution: 10 (I added this)

Amount of ounces of iodine in new solution:



How do I proceed ?

[Guest]

amount of pure iodine in 10 oz of 10% solution is 0.1 x 10 = 0.01

amount of alcohol needed added is A to make 5% solution

iodine after = iodine before

0.05(10 + A) = 0.1 x 10
0.005 + 0.05A = 0.01
0.05 A = 0.005
A = 0.1 oz of alcohol added

[Guest]

Thanks. Simpler than I thought.

[Guest]

Your calculation has mistakes in it.

0.05(10 + A) = 0.1 x 10
is correct, but it rearranges as follows:
0.5 + 0.05 A = 1
0.05 A = 0.5
A = 10 oz of alcohol added

This makes sense: if you double the weight, you halve the concentration.

Hope this helped,

R. Baber.

[Guest]

Yes, well spotted........
I thought 0.1 oz was a very small amount.......I read over the calcs. several times but didn't spot it...
Thanks

[Guest]

Thanks again.

[Guest]

The first line of my solution calculation is incorrect as well....

"amount of pure iodine in 10 oz of 10% solution is 0.1 x 10 = 0.01"

It should read........

"amount of pure iodine in 10 oz of 10% solution is 0.1 x 10 = 1.0 oz"

[Guest]

Thanks.

[leesajohnson]

You should add 0.1 alcohol.

[Guest]

You should add 0.1 alcohol.

..............what does that mean????..........explain yourself...........