Question

A 0.1064 g sample of a pesticide was decomposed by the action of sodium biphenyl. The liberated Cl- was extracted with water and titrated with 23.28 mL of 0.03337 M AgNO3 using an adsorption indicator. Assuming that the pesticide aldrin is the only source of Cl in the sample, what is the % aldrin (C12H8Cl6, 364.92 g/mol) in the sample?

Answer 1

Answer:

Percentage of an aldrin in the sample is 44.41%.

Explanation:

$Cl^-+AgNO_3\rightarrow AgCl+NO_{3}^-$

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

$0.03337 M=\frac{n}{0.02328 L}$

n = $0.03337 M\times 0.02328 L=0.0007768 mol$

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:

$\frac{1}{1}\times 0.0007768 mol=0.0007768 mol$ 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:

$\frac{0.0007768 mol}{6}=0.0001295 mol$

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:

$\frac{0.04726 g}{0.1064 g}\times 100=44.41\%$