A chemist has one solution that is 80% acid and another solution that is 30% acid. How much of the first (80%) solution is needed to make a 400 L solution that is 62% acid?
Answers
Answered by
0
let x Ltr of 80% solution is mixed with 400-x ltr of 30% solution
ATQ we have x*80/100 + (400-x)*30/100 = 400*62/100
⇒ 80x + 12000 - 30x = 24800
⇒ 50x = 12800
x = 256 ltr
ANS- 256 ltr of 80% acid solution is required
ATQ we have x*80/100 + (400-x)*30/100 = 400*62/100
⇒ 80x + 12000 - 30x = 24800
⇒ 50x = 12800
x = 256 ltr
ANS- 256 ltr of 80% acid solution is required
Similar questions