. 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?

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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?

Explanation:Let the molar mass of KBr is [tex]M_{1}[/tex] and NaBr is [tex]M_{2}[/tex].In 21.545 g, the moles of KBr and NaBr are [tex]n_{1}[/tex] and [tex]n_{2}[/tex].Therefore, [tex]M_{1} \times n_{1} + M_{2} \times n_{2}[/tex] = 21.545 g [tex]n_{2} = \frac{21.545 g - M_{1} \times n_{1}}{M_{2}}[/tex]Mass of AgBr is 35.593 g. [tex](M_{Ag} + M_{Br})(n_{1} + n_{2})[/tex] = 35.593 g [tex](M_{Ag} + M_{Br})n_{1} + (M_{Ag} + M_{Br})n_{2}[/tex] = 35.593 g [tex]M_{AgBr} + (M_{Ag} + M_{Br})\frac{21.545 g - M_{1} \times n_{1}}{M_{2}}[/tex] = 35.593 g [tex]n_{1} = \frac{35.593 g \times M_{2} - M_{AgBr} \times 21.545 g}{M_{AgBr} (M_{2} - M_{1})}[/tex]Now, putting the given values into the above formula as follows. [tex]n_{1} = \frac{35.593 g \times M_{2} - M_{AgBr} \times 21.545 g}{M_{AgBr} (M_{2} - M_{1})}[/tex] [tex]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)}[/tex] = [tex]\frac{3662.306 - 4045.552}{-1710.229} mol[/tex] = [tex]\frac{383.246}{1710.229} mol[/tex] = 0.224 molHence, mass of KBr will be calculated as follows. [tex]0.224 mol \frac{119.002 g}{1 mol}[/tex] = 26.656 gThus, we can conclude that mass of KBr present in the original mixture is 26.656 g.

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