The mass of iron (III) oxide (Fe2O3) : 85.12 g
Given
3.20x10²³ formula units
Required
The mass
Solution
1 mole = 6.02.10²³ particles
Can be formulated :
N = n x No
N = number of particles
n = mol
No = 6.02.10²³ = Avogadro's number
mol of Fe₂O₃ :
[tex]\tt n=\dfrac{3.2.10^{23}}{6.02.10^{23}}=0.532[/tex]
mass of Fe₂O₃ (MW=160 g/mol)
[tex]\tt mass=mol\times MW=0.532\times 160=85.12~g[/tex]
We have 84.84g of Fe[tex]_2[/tex]O[tex]_3[/tex] in 3.20*10^23 particles.
Data;
To solve this problem, we need to relate Avogadro number to the molar mass.
From molar mass - Avogadro's number relationship;
[tex]159.69g = 6.023*10^2^3\\xg = 3.20*10^2^3\\x = \frac{159.69*3.20*10^2^3}{6.023*10^2^3} \\x = 84.84g[/tex]
From the above calculations, 84.84g of Fe[tex]_2[/tex]O[tex]_3[/tex] is present in 3.20*10^23 particles of Fe[tex]_2[/tex]O[tex]_3[/tex].
Learn more on Avogadro's number here;
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