How many milliliters of water must be mixed with 30 milliliters of a 36% sulfuric acid solution to obtain a 23% sulfuric acid solution?
+23251 Let X = amount of water
Final Amount = X + 30
Percentage of sulfuric acid in water = 0%
Final Amount x Final Percentage = (First Amount x First Percentage) + (Second Amount x Second Percentage)
(X + 30) x (0.23) = ( 30 x 0.36 ) + ( X x 0.00 )
0.23X + 6.90 = 10.80 + 0.00
0.23X + 6.90 = 10.80
0.23X = 3.90
X = 16.96 ml
