Question

A mixture of ${\text{KBr}}$ and ${\text{NaBr}}$ weighing $0.560{\text{ g}}$ was treated with aqueous ${\text{A}}{{\text{g}}^ + }$ and all the bromide ion was recovered as $0.970{\text{ g}}$ of pure ${\text{AgBr}}$ . The fraction by weight of ${\text{KBr}}$ in the sample is:
A)$0.25$
B)$0.38$
C)$0.2378$
D)$0.285$

Answer 1

Hint:To solve this question, it is required to have knowledge about the ionic reaction occurring when silver ion is added to a solution containing bromide ion. From the weight given, we shall find the moles of ${\text{AgBr}}$ produced and the weight of bromide ion using the mole concept (formula given). Then, we shall equate it with the moles of ${\text{KBr}}$ and ${\text{NaBr}}$ to find the mass of ${\text{KBr}}$. At last, we shall calculate the fraction of ${\text{KBr}}$ in the sample.

Formula used: ${\text{moles = }}\dfrac{{{\text{mass given}}}}{{{\text{molar mass}}}}$

Complete step by step answer:
When ionic compounds containing bromide ions react with silver ions, ${\text{AgBr}}$ is formed which precipitates in the solution. The balanced equation of the reaction occurring will be:
${\text{KBr + NaBr + 2A}}{{\text{g}}^ + } \to 2{\text{AgBr}}$
The molar mass of ${\text{AgBr}}$ is $187.77 \sim 188{\text{ g mo}}{{\text{l}}^{ - 1}}$ . The moles of ${\text{AgBr}}$ in the solution will be:
${\text{moles = }}\dfrac{{{\text{0}}{\text{.97}}}}{{{\text{188}}}}$
$ \Rightarrow {\text{moles = 0}}{\text{.0051 g}}$
So now, we shall find the mass of ${\text{KBr}}$ present:
Let the mass of ${\text{KBr = x}}$ . So, according to the question the mass of ${\text{NaBr = 0}}{\text{.56 - x}}$ . Now, using mole concept, we shall find the moles of ${\text{KBr}}$ and ${\text{NaBr}}$ in terms of x. The molar mass of ${\text{KBr}}$ is $119{\text{ g mo}}{{\text{l}}^{ - 1}}$ and the molar mass of ${\text{NaBr}}$ is $103{\text{ g mo}}{{\text{l}}^{ - 1}}$ .
${\text{moles of KBr = }}\dfrac{{\text{x}}}{{119}}$
${\text{moles of NaBr = }}\dfrac{{0.56 - {\text{x}}}}{{103}}$
Now, from the equation, we know that the sum of the number of moles of ${\text{KBr}}$ and ${\text{NaBr}}$ are equal to the number of moles of ${\text{AgBr}}$ .
$ \Rightarrow {\text{moles of KBr + }}{\text{ moles of NaBr = moles of AgBr}}$
$ \Rightarrow \dfrac{x}{{119}} + \dfrac{{0.56 - x}}{{103}} = 0.0051$
Solving this for x, we get:
${\text{x = 0}}{\text{.1332 g}}$
Now, we have the mass of ${\text{KBr}}$ and the total mass of ${\text{KBr}}$ and ${\text{NaBr}}$ . So, we can calculate the mass fraction of ${\text{KBr}}$ in the sample.
$ \Rightarrow {\text{mass fraction = }}\dfrac{{0.1332}}{{0.56}}$
$ \Rightarrow {\text{mass fraction = }}0.2378$
$\therefore $ The correct option is option C, i.e. $0.2378$ .

Note: The mass fraction of any compound will always be between zero and one. It can be converted to percentage composition by multiplying it with 100. Mass fraction has no units and is dimensionless.