Solved: How many moles and numbers of ions of each type

Solution 14P

Here we have to calculate how many moles and numbers of ions of each type are present in the following aqueous solutions.

Step 1

(a) 130. mL of 0.45M aluminum chloride

1st we have to write the chemical reaction for the dissociation of AlCl3 (aluminum chloride)

AlCl3 (s) → Al3+(aq) + 3 Cl-(aq)

In the above reaction it has been found that 1 mol of Al3+ and 3 mole of Cl- is dissolved per mole of AlCl3.

In this question it has been given that molarity of AlCl3 is 0.45 M and volume is 130 mL.

Mole of Al 3+ formed from AlCl3 :-

130 mL of

= 0.059 mol of Al3+

Mole of Cl - formed from AlCl3 :-

130 mL of

= 0.176 mol of Cl-

Number of Al3+  and Cl - ions:-

Number of ions can be obtained by multiplying mole with Avogadro’s number, thus

Number of Al3+ ion =   0.059 mol of Al3+ = 0.3551023 Al3+ ions

Number of Cl- ion =   0.176 mol of Cl- = 1.061023 Cl- ions

Thus the mole of Al3+ and Cl- ion is found to be 0.059 mole and 0.176 mole & number of ions is found to be 0.3551023 Al3+ ions and 1.061023 Cl- ions.