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

Answer:

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.

$0.05 \frac{mgAs}{L} x \frac{1molAs}{74.92*10^{3}mgAs} x \frac{1molAsCl_{3}}{1molAs} x \frac{181.2gAsCl_{3}}{1molAsCl_{3}} = 1.209*10^{-4}\frac{gAsCl_{3}}{L}$

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:      

$1.209*10^{-4} \frac{gAsCl_{3}}{L} 8.2*10^{5} L = 99.2 gAsCl_{3}$.

Have a nice day!