What is the molarity of chloride ion in a solution made by dissolving 1.300g of aluminumchloride in a total volume of 500.0mL?

0.0582 M

Explanation:

The following data were obtained from the question:

Mass of AlCl3 = 1.3 g

Volume of water = 500 mL

Molarity of chloride ion (Cl¯) =?

Next, we shall determine the number of mole in 1.3 g of AlCl3. This can be obtained as follow:

Mass of AlCl3 = 1.3 g

Molar mass of AlCl3 = 27 + (35.5×3)

= 27 + 106.5

= 133.5 g/mol

Mole of AlCl3 =?

Mole = mass /molar mass

Mole of AlCl3 = 1.3/133.5

Mole of AlCl3 = 0.0097 mole

Next, we shall convert 500 mL to litres (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

500 mL = 500 mL × 1 L / 1000 mL

500 mL = 0.5 L

Thus, 500 mL is equivalent to 0.5 L.

Next, we shall determine the molarity of the AlCl3 solution. This can be obtained as follow:

Mole of AlCl3 = 0.0097 mole

Volume of water = 0.5 L

Molarity of AlCl3 =?

Molarity = mole /Volume

Molarity of AlCl3 = 0.0097 / 0.5

Molarity of AlCl3 = 0.0194 M

Next, we shall write the dissociation equation of AlCl3 in solution. This is illustrated below:

AlCl3 (aq) —> Al³⁺ (aq) + 3Cl¯ (aq)

From the balanced equation above,

1 mole of AlCl3 produced 3 moles of Cl¯.

Finally, we shall determine the molarity of the chloride ion Cl¯ in the solution as follow:

From the balanced equation above,

1 mole of AlCl3 produced 3 moles of Cl¯.

Therefore, 0.0194 M AlCl3 will produce = 0.0194 × 3 = 0.0582 M Cl¯.

Thus, the molarity of the chloride ion, Cl¯ in the solution is 0.0582 M.

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