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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?

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gautamisahoo
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

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