Question

Calculate the number of chloride ions in 100 mL of 0.01 M AlCl3 solution.

Answer 1

Answer:

• The no. of chloride ions is 1.806 × 10²¹

Given:

1. Volume (V) = 100 ml = 0.1 L
2. Molarity (M) = 0.01 M

Explanation:

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Firstly lets find the moles of Al Cl₃ in the solution.

From the formula we know,

⇒ M = n/V

Here,

• M Denotes Molarity.
• n Denotes moles.
• V Denotes volume (in liters)

Substituting the values,

⇒ 0.01 = n/ 0.1

⇒ n = 0.01 × 0.1

⇒ n = 0.001

⇒ n = 10⁻³

⇒ n = 10⁻³ moles.

∴ There are 10⁻³ moles of Al Cl₃ in the solution.

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From the Chemical reaction we know,

★ Al Cl₃ ↔ Al³⁺ + 3 Cl⁻

(Note Backward reaction is also possible)

Now, we can conclude that,

1 mole of Al Cl₃ gives 3 moles of Chloride ions. so,

⇒ 1 mole of Al Cl₃ = 3 moles of Cl⁻ ions

⇒ 10⁻³  mole of Al Cl₃ = 3 × 10⁻³ moles of Cl⁻ ions

⇒ 10⁻³  mole of Al Cl₃ = 3 × 10⁻³ moles of Cl⁻ ions

∴ In the given solution there are 3 × 10⁻³ moles of Cl⁻ ions.

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We know.

⇒ n = N₀/N

Here,

• n Denotes moles.
• N₀ Denotes atoms
• N Denotes Avogadro's number.

Substituting the values,

⇒ 3 × 10⁻³ = N₀/ 6.023 × 10²³

⇒ N₀ = 3 × 10⁻³ × 6.023 × 10²³

⇒ N₀ = 18.06 × 10⁽²³ ⁻ ³⁾

⇒ N₀ = 18.06 × 10²⁰

⇒ N₀ = 1.806 × 10²¹

⇒ N₀ = 1.806 × 10²¹

∴ The no. of chloride ions is 1.806 × 10²¹.

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