A chemist wishes to prepare 100 liters of 45% purity of sulphuric acid.He has two kinds of acids solutions in stock,one is 55% pure and other is 33% pure.How many liters of each kind should be used for the mixture?
Answers
Step-by-step explanation:
Let's begin by assigning letters to represent our two unknowns:
x (liters of 55% solution)
y (liters of 33% solution)
Our system of equations will consist of two equations:
Equation #1 (total volume of solution)
Equation #2 (total concentration of acid)
Our total volume of solution is 100 liters, which can be expressed as the sum of our unknowns:
Equation #1: x + y = 100
Our total concentration of acid can be expressed as the sum of the individual acid concentrations to make up the concentration of the final solution:
Equation #2: (0.55)(x) + (0.33)(y) =
(0.45)(100)
We can use Equation #1 to express one unknown in terms of the other and then plug that expression into Equation #2 to solve for one of the unknowns:
x + y = 100
y = 100 - x
Now we'll plug our expression for y in terms of x into Equation #2 and solve for x:
0.55(x) + 0.33(100 - x) = 0.45(100)
55(x) + 33 (y) = 4500
55x + 33 (100-x) = 4500
x = 600/11 (liters of 55% solution)
Now we'll plug our value for x into Equation #1 and solve for y:
600/11 + y = 100
y = 100 - 600/11
y = 500/11 (liters of 33% solution)
Finally, we will verify the correctness of our answers by plugging these values into Equation #2 to see if the sum of the component acid concentrations equals the final solution concentration:
0.55(600/11) + (0.33)(500/11) = 0.45(100)
30 + 15 = 45