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Relationship between Vapor Density & Relative Molecular Mass

1.24 g P is p...

Question

$1.24 g P$ is present in $2.2 g$ of :

P_{2} S_{4}

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P_{2} S_{2}

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P_{4} S_{3}

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P S_{2}

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Solution

The correct option is C $P_{4} S_{3}$
atomic weight of phosphorus $= 31$

atomic weight of sulphur $= 32$

emperical formula $= - (1.24 / 31) : (0.96 / 32)$

$= 4 : 3$

$∴$ formula is $P_{4} S_{3}$

So the option C is correct.

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Similar questions

1.24 gm P is present in 2.2 gm of

P - P

P_{4} S_{3}

P_{4} O_{10}

440

P_{4} S_{3}

384 g m

O_{2}

P_{4} S_{3} + O_{2} \to P_{4} O_{10} + S O_{2}

P_{4} S_{3} + 8 O_{2 \to} P_{4} O_{10} + 3 S O_{2}

P_{4} S_{3}

0.96

{PR}^{2} = {PS}^{2} + QR \times ST + {(\frac{QR}{2})}^{2}

{PQ}^{2} = {PS}^{2} - QR \times ST + {(\frac{QR}{2})}^{2}

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