Question

Part If 250 ml of AgNO3 is needed to precipitate all the Cllons in a 0800-mg sample of KCl forming AgCl) what is the molarity of the AgNO3 solution?

Answer 1

0.800 mg = (0.800 / 1000) g

The equation is as follows-

AgNO3 + KCl → AgCl + KNO3

according to the equation:

1mol AgNO3 reacts with 1 mol KCl

Molar mass KCl = Molar mass of Potassium + Molar mass of Chlorine ( K + Cl)

Molar mass KCl = 39.1+35.5 = 74.6g/mol

Which leads us to

(0.800/1000) / 74.6 = 1.072 x 10 ^-5 mol KCl [ mass divided by molar mass ]

We have 1.072 x 10 ^-5 mol KCl, therefore you have 1.072 x 10 ^-5 mol AgNO3 in 25.0ml

Mol AgNO3 in 1000ml = (Number of moles of AgNO3) * 1000/volume of AgNO3

Mol AgNO3 in 1000ml = (1.072 x 10 ^-5) * 1000/25.0 = 4.288 x 10 ^-4 mol AgNO3 per litre.

Molarity of AgNO3 solution = 4.288 x 10^-4 M

Cheers and Good Luck!!