Question

Question #0c0b4

Answer 1

`"DF" = 11`

Explanation:

So, you know that your final effluent contains `0.07` parts per billion of hexachlorobenzene, but that you need to get the concentration down to `"0.0065"mu"g/L"`.

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 `10^9`.

`"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 `"1 g/mL"`, you can say that 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 `10^6mu"g" = "1 g"`, you get

`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 `0.07mu"g"` of hexachlorobenzene. However, you need this value to drop to `0.0065mu"g/L"`.

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)`