Question

A certain paint mixture weighing 300 lb contains 20% solids suspended in water. How many pounds of water must be allowed to evaporate to raise the concentration of solids to 22%?

Answer 1

Answer:

27.27 lb of water must be allowed to evaporate

Explanations:

Weight of the paint mixture = 300 lb

The paint mixture contains 20% solids suspended in water

Weight of solids suspended in water = (20/100) x 300

Weight of solids suspended in water = 60 lb

Let the weight of the mixture after evaporation be w

`\begin{gathered} (22)/(100)* w\text{ = 60} \\ 0.22w\text{ = 60} \\ w\text{ = }(60)/(0.22) \\ w\text{ = }272.73\text{ lb} \end{gathered}`

The weight of the mixture after evaporation = 272.73lb

The weight of water that will be allowed to evaporate = 300 lb - 272.73 lb

The weight of water that will be allowed to evaporate = 27.27 lb

Therefore, 27.27 lb of water must be allowed to evaporate