Calculate the number of Aluminium ions present in 0.051 grams of Aluminium oxide

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6.022 \times 10^{21}

6.022 \times 10^{22}

6.022 \times 10^{20}

6.022 \times 10^{23}

answer is D.

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Detailed Solution

1 mole of

A l_{2} O_{3} = 102 g

We now that 102 g of $A l_{2} O_{3} = 6.022 \times 10^{23}$ molecules of $A l_{2} O_{3}$

The number of atoms present in given mass

$= (G i v e n m a s s \div m o l a r m a s s) \times N_{A}$

Then, 0.051 g of $A l_{2} O_{3}$ contains $= \frac{0.051}{102} \times 6.022 \times 10^{23}$ molecules of $A l_{2} O_{3}$

$= 3.011 \times 10^{20}$ molecules of $A l_{2} O_{3}$

The number of $A l^{+ 3}$ ions present in one mole of $A l_{2} O_{3}$ is 2

$∴$ the number of $A l^{+ 3}$ ions present in $3.011 \times 10^{20}$ molecules (0.051 g) of $A l_{2} O_{3}$

$= 2 \times 3.011 \times 10^{20}$

$= 6.022 \times 10^{20}$ .

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