The mass of copper atoms have the same number of atoms as there are in 4.21 g sample of silicon is 9.55 g
1 mole of silicon = 6.02×10²³ atoms
But
1 mole of silicon = 28 g
Thus, we can say that:
28 g of silicon = 6.02×10²³ atoms
28 g of silicon = 6.02×10²³ atoms
Therefore,
4.21 g of silicon = (4.21 × 6.02×10²³) / 28
4.21 g of silicon = 9.05×10²² atoms
Thus, 9.05×10²² atoms is present in 4.21 g of silicon
From Avogadro's hypothesis ,
6.02×10²³ atoms = 1 mole of copper
But
1 mole of copper = 63.55 g
Thus, we can say that:
6.02×10²³ atoms = 63.55 g of copper
Therefore,
9.05×10²² atoms = (9.05×10²² × 63.55) / 6.02×10²³
9.05×10²² atoms = 9.55 g of copper
Thus, 9.55 g of copper contains the same number of atoms in 4.21 g sample of silicon
Learn more about Avogadro's number:
https://brainly.com/question/26141731
#SPJ1
