Question

When a 1.22-g sample of pesticide was analyzed this way, it required 25.0 mL of 0.102 M Ag+ solution to precipitate all of the AsO43-. What was the mass percentage of arsenic in the pesticide? (enter the answer as a percentage, i.e. 87% would be entered as 87)

Answer 1

Answer : The mass percentage of arsenic in the pesticide is, 5.22 %

Explanation :

First we have to calculate the moles of $Ag^+$.

$\text{Moles of }Ag^+=\text{Molarity of }Ag^+\times \text{Volume of solution}=0.102mole/L\times 0.025L=0.00255mole$

Now we have to calculate the moles of arsenic.

$\text{Moles of As}=\frac{\text{Moles of }Ag^+}{3}$

$\text{Moles of As}=\frac{0.00255}{3}=0.00085mole$

Now we have to calculate the mass of arsenic.

$\text{Mass of As}=\text{Moles of As}\times \text{Molar mass of As}$

Molar mass of arsenic = 74.9 g/mole

$\text{Mass of As}=0.00085mole\times 74.9g/mole=0.0637g$

Now we have to calculate the mass percentage of arsenic in the pesticide.

$\text{Mass percent of As}=\frac{\text{Mass of As}}{\text{Total mass of sample}}\times 100$

$\text{Mass percent of As}=\frac{0.0637}{1.22}\times 100=5.22\%$

Therefore, the mass percentage of arsenic in the pesticide is, 5.22 %