Question

How many grams of${\mathbf{Br}}^{\mathbf{-}}$were in a sample that produced 1.000 g of

AgBr precipitate (FM 5 187.77)?

Short answer

The number of bromide ions generated in a sample that can cause silver bromide precipitation is 0.4255g

Step-by-step solution

Define molarity

The ratio of moles of solute to the volume of the solution is known as molarity (in litres).

Molartiv(M) = Moles of solute(inmol)

Find the mass of  AgBr

The amount of bromide ions generated in a sample that can result in silver bromide precipitation.

The number of moles of silver bromide is computed using 1.00 g of AgBr

The number of moles of silver bromide to the number of moles of bromide ion is 1 : 1

$$\begin{gathered} \mathrm{Molar}\mathrm{mass}=\frac{\mathrm{Mass}}{\mathrm{Mole}} \\ =79.904\mathrm{g}/\mathrm{mol}\times 0.00533 \\ =0.4255\mathrm{g} \end{gathered}$$

Therefore, the amount of bromide ions formed in a sample that can produce silver bromide precipitate is 0.4255 g.