Question

$5.0g$ of $KCl{O_3}$ gave $0.03mol$ of ${O_2}.$ hence, percentage purity of $KCl{O_3}$ is: $2KCl{O_3} \to 3KCl + 3{O_2}$

Answer 1

Hint: We know that percentage purity of substance or compound can be determined by dividing the mass of the pure chemical by the total mass of the sample, and then multiplying this number by hundred to get percentage purity.


Complete answer:
 In the given question it is given that $5.0g$ of $KCl{O_3}$ gave $0.03mol$ of ${O_2}.$we have to calculate percentage purity of $KCl{O_3}$ . Now calculate the percentage purity step by step:
Given mass of $KCl{O_3} = 5.0g$and we know the molar mass of $KCl{O_3}$ is $122.5g/mol.$
So, in $5.0g$ the moles of $KCl{O_3}$ is $\dfrac{5}{{122.5}}$ $ = 0.04moles$
From the given reaction it given that $2$ moles of \[KCl{O_3}\] produced $3$ moles of ${O_2}$
So, $0.04$ moles of $KCl{O_3}$ produce $\dfrac{3}{2} \times 0.04$ moles of ${O_2}$
Thus, $0.04$ moles of $KCl{O_3}$ produced $0.06$ moles of ${O_2}$
But in the question only $0.03$ moles of ${O_2}$ is given. So, the percentage purity is,
Percentage purity $ = \dfrac{{0.03}}{{0.06}} \times 100$
On solving above equation, we get
Percentage purity $ = 50\% $

The percentage purity of $KCl{O_3}$ is $50\% $ . So, the correct option is $B$ .


Additional information:Percentage purity: percentage purity is defined as it is the percentage of pure compound in an impure sample. It can also define as percentage purity indicates the amount of pure and impure substance present in a sample.


Note: It is to be noted that in an impure sample of a chemical of known percent purity is used in a chemical reaction, the percent purity has to be used in stoichiometric calculations. Conversely, the percent purity of an impure sample of a chemical of unknown percent purity can be determined by a reaction with a pure compound as in an acid-base titration.