Answer:
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.
