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