How many liters of water must be added to 35 liters of 80% hydrochloric acid solution to reduce its strength to 75%?
Ans: 6.53 liters
solution:
x (0%water) +35 (80% HCL) = (35+x)(75% strength HCL)
35*(0.8) = (35+x)0.75
x = 2.3333 liters
incorrect
How many liters of water must be added to 35 liters of 80% hydrochloric acid solution to reduce its strength to 75%?
Ans: 6.53 liters
solution:
x (0%water) +35 (80% HCL) = (35+x)(75% strength HCL)
35*(0.8) = (35+x)0.75
x = 2.3333 liters
incorrect
Hello, MathMinors!
How many liters of water must be added to 35 liters of 80% HCl solution
to reduce its strength to 75%?
Ans: 6.53 liters . . . . Wrong!
We concerned with the amount of water to be added.
The amount of water is our concern.
We start with 35 liters of soluton which is 20% water,
. . It contains:. liters of water.
We add liters of water.
The mixture contains:. liters of water. .[1]
We know that the mixture will contain liters of which 25% is water.
. . The mixture contains:. liters of water. .[2]
We just described the final amount of water in two ways.
There is our equation . . .
Solve for
. . . . . . . . . .
Using a concentration line of 0%, 75%, 80%
80 -75 = parts H2O =5 , 75-0 = parts of 80% to make 80 parts of 75%
parts of water required for 35 parts 80% = 35/75 *5 = 2.333 parts water.
Water solutions of HCL exceeding 40% do not exist at normal temp and pressure