Question

A solution of 44 liters has acid and water are in the ratio of5:6.If 4 liters of the solution is replaced with 4 liters of acid .What is the amount of acid present in the new mixture

Answer 1

sum of ratios=5+6=11

acid=5/11×44=20 L

water=6/11×44=24L

replacing taking place.........

new solution consists:

24L of acid and 20 L of water

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