Question

A 50.00-mL solution containingwas treated with excess AgNo₃ to precipitate 0.2146 g of AgBr (FM 187.772). What was the molarity of NaBr in the solution?

Short answer

The Molarity of NaBr in a solution is $\left[\mathrm{NaBr}\right]=0.02286\mathrm{M}$.

Step-by-step solution

calculate the moles of AgBr: 

In this task we have a 50 mL solution that contains NaBr treated with excess AgNO₃in order to precipitate 0.2146 g of AgBr (FM = 187.772). Here we will calculate the molarity of NaBr in the solution.

Consider this reaction:

$\mathrm{NaBr}+{\mathrm{AgNO}}_3\to \mathrm{AgBr}+{\mathrm{NaNO}}_3$

From the above reaction we can see that $\mathrm{n}\left(\mathrm{AgBr}\right)=\mathrm{n}\left(\mathrm{NaBr}\right)$.

Calculate the moles of AgBr:

$$\begin{gathered} \mathrm{n}\left(\mathrm{AgBr}\right)=\frac{\mathrm{m}}{\mathrm{M}} \\ \mathrm{n}\left(\mathrm{AgBr}\right)=\frac{0.2416\mathrm{g}}{187.772\mathrm{g}/\mathrm{mol}} \\ \mathrm{n}\left(\mathrm{AgBr}\right)=1.1429\times {10}^{-3}\mathrm{mol} \end{gathered}$$

Calculate the molarity of NaBr:

$$\begin{gathered} [\mathrm{NaBr}]=\frac{\mathrm{n}}{\mathrm{V}} \\ \left[\mathrm{NaBr}\right]=\frac{1.1429\times {10}^{-3}\mathrm{mol}}{50\times {10}^{-3}\mathrm{L}} \\ \left[\mathrm{NaBr}\right]=0.02286\mathrm{M} \end{gathered}$$