Answer:
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
Answer:
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
