Question

Leachate is produced when precipitation infiltrates a sanitary landfill,

contacts the waste material, and appears at the bottom of the stored waste. Assume 6.9 kg of

benzene

(C 6 H 6 ) were placed in the landfill and it is all dissolved in the 90,000 gallons of leachate produced

during 1 year. What is the benzene concentration in the leachate during this 1 year in (a) mg/L, (b)

ppb m and (c) moles/L?

Answer 1

Answer – Given, mass of benzene = 6.9 kg, volume = 90000 gallons

First we need to convert given mass gallons to L

We know,

1 gal = 3.785 L

90000 gal = ?

= 340650 L

a)Benzene concentration in mg/L

We need to convert the mass kg to mg
1 kg = 1.0*10⁶ mg

6.9 kg = ?

= 6.9*10⁶ mg

So, benzene concentration in mg/L = 6.9*10⁶ mg / 340650 L

                                                    = 20.25 mg/L

b) Benzene concentration in ppb and ppm

We know ppb means part per billion means microgram per liter

So, we need to convert the given mass kg to µg

1 kg = 1.0*10⁹ µg

6.9 kg =?

= 6.9*10⁹ µg

benzene concentration in ppb = 6.9*10⁹ µg / 340650 L

                                            = 20255.39 ppb

ppm means mg/L, so benzene concentration in ppm = 20.25 mg/L

c) Benzene concentration in mol /L

We need to calculate moles of benzene

Moles = given mass / molar mass

         = 6.9*10³ g / 78.11 g.mol^-1

         = 88.34 moles

So, benzene concentration in mol /L = 88.34 mole / 340650 L

                                                     = 0.000259 mol/L