Question

Practice 1. An aluminum wire is placed in 60.0 mL of an unknown copper(II) chloride dihydrate solution. The copper(II) and aluminum react in the redox reaction below, producing solid copper: 3 Cu (aq) 2 Al(s) -3 Cus) 2 Al (aq) The solid copper was filtered using a Büchner funnel, and then was dried one time and weighed to be 0.256 g. It was dried a second time and weighed to be 0.235 g. It was dried a third time and weighed to be 0.234 g. What was the original concentration in units of Molarity) of the copper(II) chloride dihydrate solution?

Answer 1

Let us consider a reaction,

3CuCl₂ (aq) + 2Al(s) ---> 3Cu(s) + 2AlCl₃ (aq)

Given data,

Mass of copper = 0.234 g

Moles of solid copper = Mass / molar mass

= 0.234 g / 63.54 g/mol

= 0.00368 mol

from the reaction,

1 mole of copper is present in 1 mole of copper chloride.

0.00368 moles comprises with 0.00368 mole copper chloride dihydrate CuCl₂ . 2H₂O

Volume of solution = 60 mL = 60 / 1000 = 0.060 L

Molarity of copper copper chloride dihydrate = Mole / Volume

= 0.00368 mole / 0.060

= 0.061 M