Calculate the number of aluminium ions present in - 0-051- g of aluminium oxide- nbsp-Atomic mass of aluminium- - 27u-

1 mole of Al_2O_3=2× 27+3× 16 #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; #160; =102 gMoles of ...

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Standard XII

Chemistry

Concentration Terms - An Introduction (w/w etc)

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Question

Calculate the number of aluminium ions present in $0.051$ g of aluminium oxide. (Atomic mass of aluminium= $27 u$ )

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Solution

$1$ mole of $A l_{2} O_{3} = 2 \times 27 + 3 \times 16$
$= 102$ g

Moles of $A l_{2} O_{3} = \frac{0.051}{102} = 5 \times 10^{- 4} m o l e s$

1 mole contains $6.02 \times 10^{23} m o l e c u l e s$

$∴ 5 \times 10^{- 4}$ g of $A l_{2} O_{3}$ contains = $5 \times 10^{- 4} \times 6.02 \times 10^{23} =$ $3.011 \times 10^{20}$ molecule

no. of $A l^{3 +}$ ions in $1$ molecules of $A l_{2} O_{3}$ is $2$ .

$∴$ No. of aluminium ions will be:-
$2 \times 3.011 \times 10^{20} = 6.022 \times 10^{20}$ .

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Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.

What is meant by chemical formula? Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.
(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u)
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Calculate the number of aluminium ions present in 0.051 g of aluminium oxide. (Atomic mass of aluminium= 27u)

1 mole of Al2O3=2×27+3×16 =102gMoles of Al2O3=0.051102=5×10−4 moles1 mole contains 6.02×1023molecules∴5×10−4g of Al2O3 contains = 5×10−4×6.02×1023=3.011×1020 molecule no. of Al3+ ions in 1 molecules of Al2O3 is 2.∴ No. of aluminium ions will be:-2×3.011×1020=6.022×1020.

Calculate the number of aluminium ions present in $0.051$ g of aluminium oxide. (Atomic mass of aluminium= $27 u$ )