The mass of 1.23 x10^24 helium atoms is 8.16 grams
Calculation
Step 1 : calculate the number of moles of He using Avogadro's constant
that is according to Avogadro's law 1 mole = 6.02 x10^23 atoms
? moles= 1.23 x10 ^24 atoms
by cross multiplication
= (1.23 x 10^ 24 atoms x 1 moles) / 6.02 x10^23 atoms =2.04 moles
Step 2 : calculate mass
mass =moles x molar mass of He
from periodic table the molar mass of He is 4.0 g/mol
moles= 2.04 moles x 4.0 g/mol = 8.16 grams
The mass of [tex]\rm 1.23\;\times\;10^2^4[/tex] atoms of He is 8.16 grams.
According to the mole concept,
6.023 [tex]\rm \times\;10^2^3[/tex] atoms = 1 mole
From the question,
[tex]\rm 1.23\;\times\;10^2^4[/tex] atoms = [tex]\rm \frac{1.23\;\times\;10^2^4}{6.023\;\times\;10^2^3}\;\times\;1[/tex] moles
[tex]\rm 1.23\;\times\;10^2^4[/tex] atoms = 2.04 moles
Mole = [tex]\rm \frac{weight}{molecular\;weight}[/tex]
weight = moles [tex]\times[/tex] molecular weight
Mass of 2.04 moles of He atoms = 2.04 [tex]\times[/tex] 4
Mass of 2.04 moles of He atoms is 8.16 grams.
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