Question

. Both KBr and NaBr are soluble ionic compounds and fully dissociate in aqueous solution. A 21.545-g mixture of KBr and NaBr is dissolved in water, then a solution of AgNO3 is added so that all of the bromine present is converted to solid AgBr. The AgBr product is dried and found to have a mass of 35.593 g. What mass of KBr was present in the original mixture?

Answer 1

Explanation:

Let the molar mass of KBr is $M_{1}$ and NaBr is $M_{2}$.

In 21.545 g, the moles of KBr and NaBr are $n_{1}$ and $n_{2}$.

Therefore,   $M_{1} \times n_{1} + M_{2} \times n_{2}$ = 21.545 g

                $n_{2} = \frac{21.545 g - M_{1} \times n_{1}}{M_{2}}$

Mass of AgBr is 35.593 g.

           $(M_{Ag} + M_{Br})(n_{1} + n_{2})$ = 35.593 g

    $(M_{Ag} + M_{Br})n_{1} + (M_{Ag} + M_{Br})n_{2}$ = 35.593 g

        $M_{AgBr} + (M_{Ag} + M_{Br})\frac{21.545 g - M_{1} \times n_{1}}{M_{2}}$ = 35.593 g

              $n_{1} = \frac{35.593 g \times M_{2} - M_{AgBr} \times 21.545 g}{M_{AgBr} (M_{2} - M_{1})}$

Now, putting the given values into the above formula as follows.

                $n_{1} = \frac{35.593 g \times M_{2} - M_{AgBr} \times 21.545 g}{M_{AgBr} (M_{2} - M_{1})}$

               $n_{1} = \frac{35.593 g \times 102.894 g/mol - 187.7722 g/mol \times 21.545 g}{187.7722 g/mol (109.894 g/mol - 119.002 g/mol)}$

                    = $\frac{3662.306 - 4045.552}{-1710.229} mol$

                    = $\frac{383.246}{1710.229} mol$

                    = 0.224 mol

Hence, mass of KBr will be calculated as follows.

                          $0.224 mol \frac{119.002 g}{1 mol}$

                          = 26.656 g

Thus, we can conclude that mass of KBr present in the original mixture is 26.656 g.