Question

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

Answer 1

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 2

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