Question

How many bromine atoms are present in 30.1 g of CH2Br2

Answer 1

There are approximately 1.43 x 10^23 bromine atoms in 30.1 g of CH2Br2.

Step-by-step explanation:

Calculate the molar mass of CH2Br2:

M(CH2Br2) = M(C) + 2M(H) + 2M(Br)

M(CH2Br2) = 12.01 g/mol + 2(1.01 g/mol) + 2(79.90 g/mol)

M(CH2Br2) = 252.82 g/mol

Calculate the number of moles of CH2Br2:

n = m/M

n = 30.1 g / 252.82 g/mol

n = 0.119 mol

Determine the number of bromine atoms:

There are 2 bromine atoms in each molecule of CH2Br2, so we can use Avogadro's number to convert moles to atoms:

N = n * N_A * 2

N = 0.119 mol * 6.022 x 10^23 atoms/mol * 2

N = 1.43 x 10^23 bromine atoms