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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)

User Kekoa
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1 Answer

2 votes

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^+* \text{Volume of solution}=0.102mole/L* 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}=(0.00255)/(3)=0.00085mole

Now we have to calculate the mass of arsenic.


\text{Mass of As}=\text{Moles of As}* \text{Molar mass of As}

Molar mass of arsenic = 74.9 g/mole


\text{Mass of As}=0.00085mole* 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}}* 100


\text{Mass percent of As}=(0.0637)/(1.22)* 100=5.22\%

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

User Cent
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