Calculate the number of bromide ions present in 3.5g of AgBr2

There are 1.57 * 10²² bromide ions present in 3.5 g of AgBr₂

Explanation:

Equation for the ionization of silver bromide, AgBr2 is given below

AgBr₂ > Ag²⁺ + 2Br⁻

1 mole of AgBr2 produces 2 moles of bromide ions

molar mass of silver bromide 268 g/mol

molar mass of bromide ion 80 g/mol

number of moles in 3.5 g of AgBr₂ = mass/molar = 3.5 / 268 = 0.013 mole

0.013 moles of AgBr₂ will produce 2 * 0.013 moles of Br⁻ = 0.026 moles

One mole of a substance contains 6.02 * 10²³ particles. Therefore, one mole of bromide ions will contain 6.02 * 10²³ ions.

Number of ions present in 0.026 moles of bromide ions = 0.026 * 6.02 * 10²³ = 1.56 * 10²² ions.

Therefore, there are 1.57 * 10²² bromide ions present in 3.5 g of AgBr₂