The formula of oxide of potassium with percentage of oxygen being $45.07 %$ is :

K_{2} O

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K O_{2}

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K_{2} O_{2}

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None of these

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Solution

The correct option is A $K O_{2}$
The formula of an oxide of potassium with the percentage of oxygen being $45.07 %$ is $K O_{2}$
Here the percentage of oxygen is $\frac{32}{71} \times 100 = 45.07$ (32 is molecular weight of oxygen and 71 is the total molecular weight of $K O_{2}$ )
So the compound is $K O_{2}$
Hence option $B$ is correct.

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Similar questions

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The formula of oxide of potassium with percentage of oxygen being 45.07% is :

The correct option is A KO2The formula of an oxide of potassium with the percentage of oxygen being 45.07% is KO2Here the percentage of oxygen is 3271×100=45.07 (32 is molecular weight of oxygen and 71 is the total molecular weight of KO2)So the compound is KO2Hence option B is correct.

The formula of oxide of potassium with percentage of oxygen being $45.07 %$ is :