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If 25.0ml of AgNO3 is needed to precipitate all the Cl ions in a 0.795mg sample...

If 25.0ml of AgNO3 is needed to precipitate all the Cl ions in a 0.795mg sample of KCl what is the molarity of the AgNo3 solution?
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Answer #1

The balanced equation:
AgNO3 + KCl → AgCl + KNO3

1mol AgNO3 reacts with 1 mol KCl to form 1 mol of AgCl.

Molar mass KCl = 39.1+35.5 = 74.6 g/mol and given mass of KCl=0.795 mg=0.795x10-3 g

Moles of KCl = 0.795x10-3 g/74.6 g/mol= 1.0656*10^-5 mol KCl

You have 1.656*10^-5 mol KCl, therefore you have 1.0656*10^-5 mol AgNO3 in 25.0ml

Mol AgNO3 in 1000ml = (1.0656*10^-5) * 1000/25.0 = 4.263*10^-4 mol AgNO3 per litre

Molarity of AgNO3 solution = 4.263*10^-4 M.

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