An oxide of sulphur contains $50$ % of sulphur in it. Its empirical formula is :

S O_{2}

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S O_{3}

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S O

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S_{2} O

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Solution

The correct option is A $S O_{2}$
The molar ratio is given as,
$n_{S} : n_{O}$
$\frac{50}{32} : \frac{50}{16}$
$= 1 : 2$
Hence, the formula is $S O_{2}$ .

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