How many grams are in 6.78 x 10^24 atoms of He?

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

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