Question

Calculate the number of aluminium ions present in 0.051g of Aluminium Oxide.

Answer 1

Given :

Mass of aluminium oxide = 0.051$\sf{g}$

To Find :

No. of aluminium ions present in the given sample$.$

SoluTion :

First we have to find no. of moles of aluminium oxide in the given sample$.$

Molar mass of $\sf{Al_2O_3}$ = 102g/mol

➝ Moles = mass/molar mass

➝ Moles = 0.051/102

➝ Moles = 0.0005

1 mole of $\sf{Al_2O_3}$ contains $\sf{N_A}$ molecules of $\sf{Al_2O_3}$.

Therefore, No. of molecules in the given sample of $\sf{Al_2O_3}$ will be $\sf{0.005N_A}$

1 molecule of $\sf{Al_2O_3}$ contains two $\sf{Al^{+3}}$ ions.

Therefore, No. of ions in $\sf{0.005N_A}$ molecules will be

$\leadsto\sf\:n=2\times 0.005N_A$

$\leadsto\sf\:n=0.01\times 6.022\times 10^{23}$

$\leadsto\boxed{\bf{n=6.022\times 10^{21}\:ions}}$

Answer 2

$\huge\mathfrak\green{answer!}$

To calculate the number of aluminium ions in 0.051g of aluminium oxide:

1 mole of aluminium oxide = 6.022 x 10²³ molecules of aluminium oxide

1 mole of aluminium oxide = 2 x Mass of aluminium + 3 x Mass of Oxygen

= (2x 27) + (3 x16) = 54 +48 = 102g

1 mole of aluminium oxide = 102g = 6.022 x 10²³ molecules of aluminium oxide

Therefore, 0.051g of aluminium oxide has = 0.051 x 6.022 x 10²³ / 102

= 3.011 x 10²⁰ molecules of aluminium oxide

One molecule of aluminium oxide has 2 aluminium ions, hence number of aluminium ions present in 0.051g of aluminium oxide = 2 x 3.011x 10²⁰ molecules of aluminium oxide

= 6.022 x 10²⁰

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