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:.$\displaystyle 0.20 \times 35 \:=\:7$ liters of water.
We add $\displaystyle x$ liters of water.
The mixture contains:.$\displaystyle x + 7$ liters of water. .[1]
We know that the mixture will contain $\displaystyle 35 + x$ liters of which 25% is water.
. . The mixture contains:.$\displaystyle 0.25(35+x)$ liters of water. .[2]
We just described the final amount of water in two ways.
There is our equation . . . $\displaystyle \boxed{x + 7 \:=\:0.25(35 + x)}$
Solve for $\displaystyle x\!:\;\;x + 7 \:=\:8.75 + 0.25x \quad\Rightarrow\quad 0.75x \:=\:1.75$
. . . . . . . . . . $\displaystyle x \:=\:\frac{1.75}{0.75} \:=\:\frac{175}{75} \:=\:\frac{7}{3}$
$\displaystyle \text{Therefore, we must add }2\tfrac{1}{3}\text{ liters of water.}$
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