Answer:
Volume of chlorine solution required is around 1.2 L
Explanation:
Volume of water in the pool = 2.0*10^4 gal
1 gal = 3.79 L
Therefore, volume of water in the pool in units of Liters would be:
[tex]\frac{2.0*10^{4}\ gal*3.79 \ L }{1\ gal} =7.6*10^{4} L[/tex]
Density of water = 1 g/ml = 1000 g/L
[tex]Mass\ of\ water\ in\ the\ pool = Density *volume\\\\=1000g/L*7.4*10^{4} L = 7.4*10^{7} g[/tex]
The accepted concentration of chlorine = 1 g/ 10⁶ g water
Therefore amount of chlorine required to disinfect the pool water would be:
[tex]=\frac{1\ g\ chlorine*7.4*10^{7}\ g water }{10^{6}\ g\ water } =74\ g[/tex]
The given solution is 6.0% w/w chlorine i.e.
6.0 g chlorine in 100 g solution
Therefore, amount of solution corresponding to 74 g chlorine would be:
[tex]=\frac{74\ g\ chlorine*100\ g\ solution}{6.0\ g\ chlorine} =1233 g[/tex]
Density of the solution = 1 g/ml
[tex]Volume\ of\ chlorine\ solution\ required = \frac{Mass}{Density}\\\\= \frac{1233g}{1.0 g/ml} = 1.2*10^{3} ml = 1.2\ L[/tex]
