Answer:
Percentage of an aldrin in the sample is 44.41%.
Explanation:
[tex]Cl^-+AgNO_3\rightarrow AgCl+NO_{3}^-[/tex]
Molarity of the silver nitrate solution = 0.03337 M
Volume of the silver nitrate = 23.28 mL = 0.02328 L
Moles of silver nitrate = n
[tex]0.03337 M=\frac{n}{0.02328 L}[/tex]
n = [tex]0.03337 M\times 0.02328 L=0.0007768 mol[/tex]
According to reaction 1 mole of silver nitrate recats with 1 moles of chloride ions.
Then 0.0007768 moles of silver nitrate will react with:
[tex]\frac{1}{1}\times 0.0007768 mol=0.0007768 mol[/tex] chloride ions.
In one mole of aldrin there are 6 moles of chloride ions.
Then moles of aldrin containing 0.0007768 moles chloride ions are:
[tex]\frac{0.0007768 mol}{6}=0.0001295 mol[/tex]
Moles of aldrin present in the sample = 0.0001295 mol
Mass of 0.0001295 moles of aldrin present in the sample :
0.0001295 mol × 364.92 g/mol =0.04726 g
Percentage of an aldrin in the sample:
[tex]\frac{0.04726 g}{0.1064 g}\times 100=44.41\%[/tex]
