Chemistry, asked by indrajitsinghvns1990, 9 months ago

1. Calcate the number of atum present in 5.1 g of allimminum oxide.​

Answers

Answered by rahulsrivastava18052
0

Answer:

The total number of atoms of oxygen present in 5.1 g aluminium oxide is 2.65 * 10²².

Explanation:

The molecular formula of aluminium oxide is Al₂O₃.

So, the molecular mass of Al₂O₃, M = 2*(27) + 3*(16) = 54 + 48 = 102 g/mol

Mass of Al₂O₃, m = 5.1 g

∴ No. of moles of Al₂O₃, n = \frac{m}{M} = \frac{1.5}{102}

M

m

=

102

1.5

= 0.0147 moles

From the formula of Al₂O₃, we can see that

In 1 mole of Al₂O₃, 3 moles of oxygen are present

∴ 0.0147 moles of Al₂O₃ will consist of = 3 * 0.0147 = 0.0441 moles of oxygen

Now, we know

1 mole = 6.022 * 10²³ atoms

∴ 0.0441 moles of oxygen = 0.0441 * 6.022 * 10²³ = 0.265 * 10²³ = 2.65 * 10²² atoms

Thus, the total number of atoms of oxygen present in 5.1 g aluminium oxide is 2.65 * 10²² atoms.

Answered by ayushi4192
0

Explanation:

The total number of atoms of oxygen present in 5.1 g aluminium oxide is 2.65 * 10²². Explanation: The molecular formula of aluminium oxide is Al₂O₃. Thus, the total number of atoms of oxygen present in 5.1 g aluminium oxide is 2.65 * 10²² atoms.

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