Question

The Greater Vancouver Regional District (GVRD) chlorinates the water supply of the region at the rate of 1 ppm, that is, 1 kilogram of chlorine per million kilograms of water. The chlorine is introduced in the form of sodium hypochlorite, which is 47.62% chlorine. The population of the GVRD is 1.8 million persons. If each person uses 750 L of water per day, how many kilograms of sodium hypochlorite must be added to the water supply each week to produce the required chlorine level of 1 ppm?

Answer 1

2835 kilograms of sodium hypochlorite must be added to the water supply each week to produce the required chlorine level of 1 ppm.

Step-by-step explanation:

Volume of water used by 1 person = 750 L

Volume of water used by 1.8 million persons : V

`V=1.8* 10^6* 750 L=1.35* 10^(9) L`

Density of water,d = 1 kg/L

Mass of water used by 1.8 million persons = m

`m=d* V=1 kg/L* 1.35* 10^(9) L=1.35* 10^(9) kg`

1 kilogram of chlorine per million kilograms of water. (Given)

Concentration of chlorine in water = 1 kg/ 1000,000 kg of water

In 1000,000 kg of water = 1 kg of chlorine

Then
`1.35* 10^(9) kg` of water have x mass of chlorine:

`\frac{x}{1.35* 10^(9) \text{kg of water}}=\frac{1 kg}{1000,000 \text{kg of water}}`

`x=(1)/(1000,000)* 1.35* 10^(9) kg=1.35* 10^3 kg`

Mass of chlorine in water of mass
`1.35* 10^(9) kg=1.35* 10^(3) kg`

Percentage of chlorine in hypochlorite = 47.62%

`47.62\%=\frac{1.35* 10^(3) kg}{\text{Total mass of sodium hypochlorite}}* 100`

Total mass of sodium hypochlorite =
`2834.94 kg\approx 2835 kg`

2835 kilograms of sodium hypochlorite must be added to the water supply each week to produce the required chlorine level of 1 ppm.