A current of 6 amperes is passed through AlCl3 solution for 15 minutes

1.	A current of 6 amperes is passed through AlCl3 solution for 15 minutes using Pt electrodes, when 0.504 g Al is produced. What is the molar mass of Al ?
Answer» Given : Electric current = I = 6 A Time = t = 15 min = 15 × 60 s = 900 s Mass of Al produced = 0.504 g Molar mass of Al = ? Reduction half reaction, \(Al^{3+}_{(aq)}\) + 3e^- ⟶Al_(aq) Quantity of electricity passed = Q = I × t = 6 × 900 = 5400 C Number of moles of electrons = \(\frac{Q}{F}\) = \(\frac{5400}{96500}\) = 0.05596 mol From half reaction, ∵ 3 moles of electrons deposit 1 mole Al ∴ 0.05596 moles of electrons will deposit, \(\frac{0.05596}{3}\) = 0.01865 mol Al Now, ∵ 0.01865 mole Al weighs 0.504 g ∴ 1 mole Al will weigh, \(\frac{0.504}{0.01865}\) = 27 g Hence molar mass of Al is 27 g mol^-1 ∴ Molar mass of Al = 27 g mol^-1

A current of 6 amperes is passed through AlCl3 solution for 15 minutes using Pt electrodes, when 0.504 g Al is produced. What is the molar mass of Al ?

Answer»

Given :

Electric current = I = 6 A

Time = t = 15 min = 15 × 60 s = 900 s

Mass of Al produced = 0.504 g

Molar mass of Al = ?

Reduction half reaction,

\(Al^{3+}_{(aq)}\) + 3e^- ⟶Al_(aq)

Quantity of electricity passed = Q = I × t

= 6 × 900

= 5400 C

Number of moles of electrons = \(\frac{Q}{F}\) = \(\frac{5400}{96500}\)

= 0.05596 mol

From half reaction,

∵ 3 moles of electrons deposit 1 mole Al

∴ 0.05596 moles of electrons will deposit,

\(\frac{0.05596}{3}\)

= 0.01865 mol Al

Now,

∵ 0.01865 mole Al weighs 0.504 g

∴ 1 mole Al will weigh,

\(\frac{0.504}{0.01865}\) = 27 g

Hence molar mass of Al is 27 g mol^-1

∴ Molar mass of Al = 27 g mol^-1

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