Question

Arnold needs a 25% solution of nitric acid. He has 20 ml of a 30% solution. How many ml of a 15% solution should he add to obtain the required 25% solution?
A. 20 ml
B. 15 ml
C. 25 ml
D. 10 ml

Answer 1

@jim_thompson5910

Answer 2

what should i do 1st

Answer 3

let x = amount of the 15% solution

Answer 4

you should always start with what you know

Answer 5

and declare what you don't and try to work towards getting it

Answer 6

He has 20 ml of a 30% solution

so he has 20*0.3 = 6 mL of pure acid

Answer 7

he also has x mL of the 15% solution, so he also has an additional 0.15x mL of acid

Answer 8

in total, he has 0.15x + 6 milliliters of pure acid

Answer 9

he has 20 mL of the 30% solution
he has x mL of the 15% solution

so he has 20+x mL of solution total

Answer 10

he wants 25% of this final solution (20+x mL) to be pure acid, so he wants 0.25(20+x) = 5 + 0.25x mL of pure acid

Answer 11

if you go back, you'll see that he has 0.15x+6 mL of pure acid, so we can say

0.15x + 6 = 5 + 0.25x

Answer 12

solve that for x to get your answer

Answer 13

D!

Answer 14

yep

Answer 15

once you get the setup down, it's not so bad