Question

Suppose 6.88 g of iron(II) chloride is dissolved in 150. mL of a 0.30 M aqueous solution of silver nitrate. Calculate the final molarity of chloride anion in the solution. You can assume the volume of the solution doesn't change when the iron(II) chloride is dissolved in it. Be sure your answer has the correct number of significant

digits.

Answer 1

The answer is quite simple

The formula for iron (Ii) chloride is FeCl₂

Which means 1 mole of iron (ll) chloride has 2 moles of chloride ions

The no of moles of iron (ll) chloride

Moles = weight /molar mass

= 6.88/126.751

= 0.054 moles

So chloride ions will be twice that

= 0.108 moles

Also we have silver nitrate in solution which will react with chloride ions as follows

So one mole of silver chloride react with 2 moles of AgNO3

Moles of AgNO₃

Will be

Since 1M 1000 ml solution has 2 mole

So 0.30 molar 150 ml solution will have

0.30×1× 150/1000

= 0.045 moles

And it will consume

Half moles of FeCl2

Which will be 0.0225 moles

Remaining moles of FeCl2

= 0.054- 0.0225

= 0.0315 moles

And chloride ions will be twice of that

= 0.0315× 2

= 0.063 moles

Molarity

= no of moles of solute /volume of sol. In liter

= 0.063/0.150

= 0.42 molar

I hope this helps if you have any query or want more detailed explanation feel free to ask in the comments section below.