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

Percentage of an aldrin in the sample is 44.41%.

Step-by-step 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=(n)/(0.02328 L)`

n =
`0.03337 M* 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:

`(1)/(1)* 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:

`(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:

`(0.04726 g)/(0.1064 g)* 100=44.41\%`