How many ml of a 78% acid solution must be mixed with 12ml of a 10% acid solution to make 30% solution?

Approximately 0.6 ml of 78% acid solution has to be mixed with 12ml of a 10% acid solution to make 30% solutionSolution:First, set up a table. Fill in the unknowns with variable "x""x" ml of 78% acid solution must be mixed with 12 ml of 10 % acid solution to make 30% acid solutionLet us solve for "x"The table is attached belowFrom the table below, we can set up two equationsSum of values of two salts = Value of mixture Sum of values of two salts = 0.78x + 1.2 Value of mixture = 0.3(x + 12) Sum of values of two salts = Value of mixture  Then, 0.78x + 1.2 = 0.3(x + 12)  0.78x + 1.2 = 0.3x + 3.6  0.78x – 0.36x = 3.6 – 1.2  4.2x = 2.4  x = 0.5714  ≈ 0.6Hence, approximately 0.6 ml of 78% acid solution has to be mixed