Question

How many mL of 1.5% (w/v) KCl solution would be needed to precipitate all of the silver as AgCl in a 5.0 g ore sample that contains 1.0% silver? Allow for a 50% excess of the KCl solution.

Answer 1

ore sample contains 1.0% silver in 5 g sample

mass of silver=> 0.01*5g=> 0.05 g

Atomic mass of silver=> 108 g/mol

so, number of moles of silver=> 0.05g/108g/mol=> 4.6X10^-4 moles

Allowing for a 50% excess, moles of Kcl required=> 1.5* 4.6X 10^-4 moles

=> 6.9 X 10^-4moles

Molecular weight of KCl=> molder mass of potassium+molar mass of chlorine => 39+35.5=> 74.5 g/mole

So, KCl required=> 6.9 X 10^-4 moles * 74.5 g/mol=> 0.052 grams

Now, per ml of 1.5% w/v solution of kcl contains=> 1.5g/100ml=> 0.015 g/ml

Therefore, volume of KCl solution required=> 0.052 g/0.015g/ml=> 3.5 ml