a lab assistant wants to make 5 liters of 27.8% acid solutions. if solutions of 50% and 13% are in stock, how many liters of each must be mixed to prepare the solution?
just work with the amount of acid in each part. If there's x L of 50% acid,
.50x + .13(5-x) = .278(5)
x=2
yup
To determine how many liters of each solution should be mixed, we can set up a system of equations.
Let's assume x represents the number of liters of the 50% acid solution and y represents the number of liters of the 13% acid solution.
Given that the total volume of the solution is 5 liters, we can express this as:
x + y = 5 (equation 1)
Now, let's consider the acid concentration. The 50% acid solution is made up of 50% acid, while the 13% acid solution is made up of 13% acid. Therefore, the amount of acid in x liters of the 50% solution is 0.5x, and the amount of acid in y liters of the 13% solution is 0.13y.
To achieve a 27.8% acid concentration in the final mixture, the amount of acid in the mixture should be 27.8% of the total volume, which is 5 liters:
0.5x + 0.13y = 0.278 * 5 (equation 2)
Now, we have a system of equations consisting of equation 1 and equation 2. We can solve this system to find the values of x and y.
Let's solve the system:
x + y = 5 (equation 1)
0.5x + 0.13y = 0.278 * 5 (equation 2)
To make the calculations more manageable, we can multiply both sides of equation 1 by 0.13 to eliminate decimals:
0.13x + 0.13y = 0.13 * 5
Simplifying equation 1:
0.13x + 0.13y = 0.65
Now, subtract equation 1 from equation 2:
(0.5x + 0.13y) - (0.13x + 0.13y) = (0.278 * 5) - 0.65
Simplifying:
0.5x - 0.13x = 1.39 - 0.65
0.37x = 0.74
x = 0.74 / 0.37
x = 2
Now, substitute the value of x into equation 1:
2 + y = 5
y = 5 - 2
y = 3
Therefore, the lab assistant should mix 2 liters of the 50% acid solution with 3 liters of the 13% acid solution to prepare the 5 liters of 27.8% acid solution.