Question

Chlorine is used to disinfect swimming pools. The accepted concentration for this purpose is 1.00 ppm chlorine, or 1.00 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 5.50 percent chlorine by mass and there are 6.52 × 104 gallons (gal) of water in the pool (1 gal = 3.79 L; density of liquids = 1.00 g/mL). Enter your answer in scientific notation.

Answer 1

4.45 L

Step-by-step explanation:

The concentration of chlorine in the pool must be 1.00g/10⁶g, and the solution has 5.50g/100g, so it will be diluted in the pool.

The product of the concentration by the volume is always constant in a solution, thus, we can use the equation:

C1V1 = C2V2

Where C is the concentration, V is the volume, 1 represents the solution, and 2 the pool. So, C1 = 5.50g/100g, C2 = 1.00g/10⁶g, V2 = 6.52x10⁴ gal = 247108 L. Thus,

(5.50/100)*V1 = (1.00/10⁶)*247108

0.055V1 = 0.247108

V1 = 4.5 L

Answer 2

The volume of a chlorine solution a homeowner should add to her swimming pool is
`4.480* 10^3 mL`.

Step-by-step explanation:

The solution contains 5.50 percent chlorine by mass.

Mass of the chlorine required to added in the pool= x

Mass of the water in the pool = y

Volume of the water ,V=
`6.52* 10^4 gal=2.464* 10^5 L`

(1 gal = 3.78 L)

V =
`2.464* 10^5 L=2.464* 10^8 mL`

( 1 L = 1000 mL)

Density of the water ,d= 1.00 g/mL

`y=d* V=1.00 g/mL* 2.464* 10^8 mL=2.464* 10^8 g`

y =
`2.464* 10^8 g`

1.00 g of chlorine per million grams of water .

Then for
`2.464* 10^8 g` of water, we required x amount of chlorine:

`z=(2.464* 10^8 g)/(10^6 g)=246.4 g`

x = 246.4 grams of chlorine

Mass of the chlorine solution = M'

`(w/w)\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}* 100`

`5.5\%=(z)/(M')* 100`

`M'=(246.4)/(5.5)* 100=4,480 g`

Density of the chlorine solution = D = 1 g/mL

Volume of the chlorine solution to be added in the pool : v

`v=D* M'=1 g/mL* 4,480 g=4480 mL`

`v=4.480* 10^3 mL`

The volume of a chlorine solution a homeowner should add to her swimming pool is
`4.480* 10^3 mL`.