Question

Find the charge of 27 g of Al3+ ions in coulombs. ??

Answer 1

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$\implies$Atomic mass of Al = 27

$\implies$ No. of moles Al =27/27

$\implies$= 1 mole

$\implies$ Charge on 1 Al3+ ion = 3 e

$\implies$ 1 e = 1.602×10^19Coulombs

$\implies$= 3 × 1.602×10^19Coulombs

$\implies$Now, Charge on 1 mole Al3+ ions


$\implies$= 3 × 1.602×10^19 × 6.022×10^23Coulombs


$\implies$= 2.894×105 Coulombs.

Answer 2

Heya ____

Solution is given here _____

Al = 27 g
Moles of Al = 27/27 = 1 mole

Charge of Al = 1Al3 + ion = 3e

1 e = 1.602 * 10^19 coulombs

3 * 1.602 * 10^19

Charge of 1 mole .....

3* 1.602 * 10^19 * 6.022* 10^23

= 2.894 * 105 ... coulombs...

Thank you