A current of 6 amperes is passed through `A1C1_(3)` solution for 15 minutes using Pt electrodes , when 0.50 g A1 is produced. What is the molar mass of A1 ?

Answer

Molar mass of A1 = 27 `g mol^(-1)`

Explanation

Given : Electric current = I = 6 A
Time = t= 15 min = 155 `xx 60 s = 900 s`
Mass of A1 produced = 0.504 g
Molar mass of A1 = ?
Reduction half reaction
`A1_((ag))^(3) + 3e^(-) to A1_((aq))`
Quantity of electricity passed = Q = `I xx t`
`=6xx 900 = 5400 C`
Number a of moles of electrons `= (Q)/(F ) = (5400)/( 96500) = 0.05596 mol`
From half reaction
:' 3 moles of electrons deposit 1 mole A1
`:. 0.05596 ` moles of electrons will deposit
`(0.05596)/(3) = 0.01865 mol A1 `
Now
`:' 0.01865 ` mol A1 weighs 0.504 g
`:. 1 ` mole A1 will weigh ,` (0.504)/(0.01865) = 27 g`
Hence molar mass of A1 is 27 g `mol^(-1)`