Question

Chlorine is used to disinfect swimming pools. The accepted concentration for this purpose is 1 ppm chlorine, or 1 g of chlorine per million grams of water. Calculate the volume of a chlorine solution (in milliliters) a homeowner should add to her swimming pool if the solution contains 6.0 percent chlorine by mass and there are 2.0 x 10^4 gallons (gal) of water in the pool (1 gal =3.79 L; density of liquids= 1.0 g/mL).

Answer 1

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:

$\frac{2.0*10^{4}\ gal*3.79 \ L }{1\ gal} =7.6*10^{4} L$

Density of water = 1 g/ml = 1000 g/L

$Mass\ of\ water\ in\ the\ pool = Density *volume\\\\=1000g/L*7.4*10^{4} L = 7.4*10^{7} g$

The accepted concentration of chlorine = 1 g/ 10⁶ g water

Therefore amount of chlorine required to disinfect the pool water would be:

$=\frac{1\ g\ chlorine*7.4*10^{7}\ g water }{10^{6}\ g\ water } =74\ g$

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:

$=\frac{74\ g\ chlorine*100\ g\ solution}{6.0\ g\ chlorine} =1233 g$

Density of the solution = 1 g/ml

$Volume\ of\ chlorine\ solution\ required = \frac{Mass}{Density}\\\\= \frac{1233g}{1.0 g/ml} = 1.2*10^{3} ml = 1.2\ L$