Respuesta :
Given:
A 76,900-liter swimming pool has a chlorine concentration of 2.6 ppm.
To find:
How many liters must be replaced with 20 ppm chlorine solution to increase the chlorine concentration to 4 ppm?
Solution:
The concentration in terms of ppm is given by:
[tex]ppm=\frac{mass\text{ of solute}}{mass\text{ of solution}}\times10^6[/tex]So, the mass of solute becomes:
[tex]mass\text{ of solute}=\frac{mass\text{ of solution}\times ppm}{10^6}[/tex]Let x mL of the pool solution is replaced by 20 ppm chlorine solution to make final chlorine concentration equal to 4 ppm. So,
[tex]\text{ mass of the solute added}+\text{ mass of the solute unchanged}=new\text{ mass of the solute in the solution}[/tex][tex]\begin{gathered} \frac{x\times20}{10^6}+\frac{(76900000-x)2.6}{10^6}=\frac{76900000(4)}{10^6} \\ 20x+199940000-2.6x=307600000 \\ 17.4x=107660000 \\ x=\frac{107660000}{17.6} \\ x=6117045.455\text{ mL} \\ x=6117.1L \end{gathered}[/tex]Thus, the answer is 6117.1 L.
