# calculate the no. of aluminium ions present in 0.051g of aluminium oxide

• 11

Aluminium oxide is Al2O3

Molar mass Al2O3 = 101.9614 g/mol
Mol Al2O3 in 0.051g = 0.051/101.9614 = 0.0005mol Al2O3
1mol Al2O3 contains 2mol Al.
Mol Al = 0.0005*2 = 0.001mol Al
1mol Al3+ contains 6.022*10^23 ions.
0.001mol contains: 6.022*10^23*0.001 = 6.022*10^20 ions
• -7

1 mole of aluminium oxide (Al2O3) = 2 � 27 + 3 � 16

= 102 g

i.e., 102 g of Al2O3 = 6.022 � 1023 molecules of Al2O3

Then, 0.051 g of Al2O3 contains =

= 3.011 � 1020 molecules of Al2O3

The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.

Therefore, the number of aluminium ions (Al3+) present in 3.011 � 1020 molecules (0.051 g ) of aluminium oxide (Al2O3) = 2 � 3.011 � 1020

= 6.022 � 1020

• -7

1 mole of aluminium oxide (Al2O3) = 2*27 + 3 * 16

= 102 g

i.e., 102 g of Al2O3 = 6.022 * 1023 molecules of Al2O3

Then, 0.051 g of Al2O3 contains =

= 3.011 * 1020 molecules of Al2O3

The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.

Therefore, the number of aluminium ions (Al3+) present in 3.011 * 1020 molecules (0.051 g ) of aluminium oxide (Al2O3) = 2 * 3.011 * 1020

= 6.022 * 1020

• 80

Given mass = 0.051 g.

Let the given mass be 'm '

Formula of compound 'Aluminium Oxide ' = Al2 O3 .

Formula mass = 102 u ( Al = 27 u and O = 16 u.)

Let the Formula mass be 'M '

Let the number of Aluminium ions present be 'n '.

Let Avagadro no. be 'n0 ' = 6.022 * 1023.

Then n = ( m* n0) / M

= ( 0.051 * 6.022 * 1023 ) / 102

= 0.307122 * 1023 / 102

= 0.003011 * 1023 ions.

= 3.011 * 1021 ions.

• 3
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