# Question #0c0b4

##### 1 Answer

#### Explanation:

So, you know that your final effluent contains **parts per billion** of hexachlorobenzene, but that you need to get the concentration down to

A concentration of **one part per billion** is equivalent to **one gram** of solute, in your case hexachlorobenzene, in **one billion grams** of solvent, in your case water.

To find the concentration of the solute in *ppb*, simply divide the mass of the solute in grams by the mass of the solvent in grams, and multiply the result by

#"ppb" = m_"solute"/m_"solvent" xx 10^9#

So, what will a **0.07 ppb** concentration mean in terms of *micrograms per liter*?

If you assume the density of water to be equal to **one liter** of water will contain

#m_"solute" = ("ppb" xx m_"solvent")/10^9#

#m_"solute" = (0.07 xx "1000 g")/10^9 = 7 * 10^(-8)"g"#

Since you know that

#7 * 10^(-8)color(red)(cancel(color(black)("g"))) * (10^6mu"g")/(1color(red)(cancel(color(black)("g")))) = 7 * 10^(-2)mu"g" = 0.07mu"g"#

So, **one liter** of contaminated water contains

How many liters of water do you need to add *per liter* of effluent to get it to reach the accepted sttandard for hexachlorobenzene?

#0.07color(red)(cancel(color(black)(mu"g"))) * "1 L"/(0.0065color(red)(cancel(color(black)(mu"g")))) = "10.77 L"#

This means that you need to add, **per liter of effluent**, a total of

#V_"water" = V_"final" - V_"effluent"#

#V_"water" = 10.77 - 1 = "9.77 L"#

of water. The dilution factor, which is defined as the final volume by the initial volume, will be

#"DF" = (10.77color(red)(cancel(color(black)("L"))))/(1color(red)(cancel(color(black)("L")))) = 10.8#

I'll leave the answer rounded to two sig figs, despite the fact that you only gave one sig fig for the ppb concentration

#"DF" = color(green)(11)#