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Chemistry
Kyra
14 January, 00:38
How many grams are in 6.78 x 10^24 atoms of He?
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Anabelle Goodman
14 January, 03:47
0
45.05g
Explanation:
Based on the Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02x10^23 atoms.
From the above findings,
1 mole of He also contains 6.02x10^23 atoms.
1 mole of He = 4g
If 4g of He contains 6.02x10^23 atoms,
then Xg of He will contain 6.78x10^24 atoms i. e
Xg of He = (4x6.78x10^24) / 6.02x10^23
Xg of He = 45.05g
Therefore, 45.05g of He contains 6.78x10^24 atoms
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Focker
14 January, 03:52
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45.05grams
Explanation:
Number of atoms present in a substance = number of moles of the substance * Avogadro's number
Given, number of atoms = 6.78*10^24
Number of moles = mass / molar mass
Molar mass of Helium = 4g/mol
Avogadro's number = 6.02*10^23
Therefore,
6.78*10^24 = (m/4) * (6.02*10^23)
6.78*10^24 = (6.02*10^23*m) / 4
Cross multiply
6.02*10^23*m=4*6.78*10^24
Divide both sides by 6.02*10^23
m = (4*6.78*10^24) / 6.02*10^23
m = (27.12*10^24) / 6.02*10^23
m=45.05grams
Therefore, there are 45.05grams in 6.78*10^24 atoms of helium
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