How many bromine atoms are present in 39.0 g of ch2br2?

Answer is: there is 2,69·10²³ atoms of bromine.

m (CH₂Br₂) = 39,0 g.

n (CH₂Br₂) = m (CH₂Br₂) : M (CH₂Br₂).

n (CH₂Br₂) = 39 g : 173,83 g/mol.

n (CH₂Br₂) = 0,224 mol.

In one molecule of CH₂Br₂, there is two bromine atoms, so:

n (CH₂Br₂) : n (Br) = 1 : 2.

n (Br) = 0,448 mol.

N (Br) = n (Br) · Na.

N (Br) = 0,448 mol · 6,022·10²³ 1/mol.

n (Br) = 2,69·10²³.