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5.0g of KClO3 gave 0.03mol of O2. hence, percentage purity of KClO3 is: 2KClO33KCl+3O2

Answer
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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 KClO3 gave 0.03mol of O2.we have to calculate percentage purity of KClO3 . Now calculate the percentage purity step by step:
Given mass of KClO3=5.0gand we know the molar mass of KClO3 is 122.5g/mol.
So, in 5.0g the moles of KClO3 is 5122.5 =0.04moles
From the given reaction it given that 2 moles of KClO3 produced 3 moles of O2
So, 0.04 moles of KClO3 produce 32×0.04 moles of O2
Thus, 0.04 moles of KClO3 produced 0.06 moles of O2
But in the question only 0.03 moles of O2 is given. So, the percentage purity is,
Percentage purity =0.030.06×100
On solving above equation, we get
Percentage purity =50%

The percentage purity of KClO3 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.
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