Question

The maximum contamination level of arsenic ion in a water system is 0.050 parts per million. If the arsenic is present as AsCl3, how many grams of arsenic chloride could be present in a system that contains 8.2 x 10^5 Liters?

Answer 1

mass AsCl₃ in 8.2x10⁵ L = 99.2gAsCl₃

Step-by-step explanation:

Let's remember that 0.05 ppm (parts per million) would be write terms of mg/L. We have 0.05 mg/L, that is a concentration unit.

First of all we need to find the g/L of AsCl₃ present in water.

`0.05 (mgAs)/(L) x (1molAs)/(74.92*10^(3)mgAs) x (1molAsCl_(3))/(1molAs) x (181.2gAsCl_(3))/(1molAsCl_(3)) = 1.209*10^(-4)(gAsCl_(3))/(L)`

Now, we just need to multiply it by 8.2 x 10⁵ L to determine the mass of AsCl₃ present in that volume.

Finally we have:

`1.209*10^(-4) (gAsCl_(3))/(L) 8.2*10^(5) L = 99.2 gAsCl_(3)`.

Have a nice day!