Question

Calculate the number of alumunium ions present in 0.051 g of Oxide.​

Answer 1

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$\Large\underline{\underline{\red{\sf \red{\maltese}\:\: Qᴜᴇsᴛɪᴏɴ :- }}}$

$\sf \small \: \text{Calculate the number of alumunium ions present in 0.051 g} \\ \sf \text{ of Oxide. \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }$

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$\Large\underline{\underline{\red{\sf \red{\maltese}\:\: Aɴsᴡᴇʀ:- }}}$

$\sf \small{Molar \: Mass \: Of \: Al_{2}O_{3} = 102 \: g}$

$\sf \small{ Al_{2}O_{3}↔2 Al^{ 3 + } + 3O ^{2 - } }$

$\sf \small{ \rightarrow102 \: g \: of \: Al_{2}O_{3} \: contains \: = 2 \times 6.02 \times 10 ^{23} \: \: Al_{3 +}ions \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \\ \sf \small { \rightarrow0.051 \: g \: of \: Al_{2}O_{3} \: contains \: = \frac{2 \times 6.02 \times 10^{23} \: \:Al ^{3 + } ions }{102} \times 0.051 } \\ \\ \sf \small \red{ Fɪɴᴀʟ \: Aɴsᴡᴇʀ= 2 \times 6.02 \times 10^{23} \: Al^{3 + } \: ions \: \: \:}$

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Answer 2

Given:

0.051 gm of Oxide.

To find:

The number of alumunium ions present in 0.051 g of Oxide.

Solution:

Oxide of alumunium = Al2O3

Now we need to calculate the weight of 1 mole of aluminum oxide :

= 2×27 + 3×16

= 54 + 48

= 102 gm

The number of molecules present in 102 gm of alumunium oxide = 6.022 × 10^(23)

So the number of molecules in 0.051 gm of iron will be:

= (0.051÷102) × 6.022 ×10^(23)

= 3.011 × 10^(20)

Now as we know that number of ions of aluminum i.e. Al^(3+) is 2 in Al2O3. Therefore number if alumunium ions present in 0.051 gm of alumunium oxide will be:

= 2× 3.011 × 10^(20)

= 6.022 × 10^(20) ions

The answer is 6.022 × 10^(20) ions.