A0.1064 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?

Percentage of an aldrin in the sample is 44.41%.

Explanation:

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

n =

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:

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:

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:

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