Answer:
0.0584 M
Explanation:
From the question given above, the following data were obtained:
Mass of CaCl₂ = 0.808 g
Volume of water = 250 mL
Concentration of chloride =?
Next, we shall determine the number of mole in 0.808 g of CaCl₂. This can be obtained as follow:
Mass of CaCl₂ = 0.808 g
Molar mass of CaCl₂ = 40 + (35.5 × 2)
= 40 + 71
= 111 g/mol
Mole of CaCl₂ =?
Mole = mass / Molar mass
Mole of CaCl₂ = 0.808 / 111
Mole of CaCl₂ = 0.0073 mole
Next, we shall convert 250 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
250 mL = 250 mL × 1 L / 1000 mL
250 mL = 0.25 L
Next, we shall determine the molarity of CaCl₂. This can be obtained as follow:
Mole of CaCl₂ = 0.0073 mole
Volume of water = 0.25 L
Molarity of CaCl₂ =?
Molarity = mole /Volume
Molarity of CaCl₂ = 0.0073 / 0.25
Molarity of CaCl₂ = 0.0292 M
Finally, we shall determine the concentration of the chloride as illustrated below:
CaCl₂ <=> Ca²⁺ + 2Cl¯
From the equation above,
1 mole of CaCl₂ produced 2 mole of Cl¯.
Therefore, 0.0292 M CaCl₂ will produce = 0.0292 × 2 = 0.0584 M Cl¯.
Thus, the concentration of the chloride ion (Cl¯) in the solution is 0.0584 M
