Question 773348: 21 gram mixture of water and potassium is .8% potassium. How much water must be evaporated so that the mixture is 1.4% potassium? Answer by Cromlix(4381) (Show Source):
You can put this solution on YOUR website! To find initial amount of potassium
x/21 * 100 = 0.8
x = 21 * 0.8/100
x = 0.168g
To find quantity of water
required to be evaporated to
attain 1.4% K
0.168/x * 100 = 1.4
1.4/100 = 0.168/x
Cross multiply.
x = 100 * 0.168/1.4
x = 12g
Therefore 9g of water must be evaporated.
Hope this helps.
:-)