A solid element burns completely in oxygen to give gaseous oxide without any change in volume. If the vapour density of oxide is $32$ , the equivalent weight of element would be?

8

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16

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24

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32

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Solution

The correct option is A $8$
$2 M + O_{2} \to 2 M O$
Molecular weight of gaseous oxide $= 2 \times$ vapour density
$= 2 \times 32 = 64$
$∴$ So molar mass of $M = 64 - 2 \times 16 = 32$
So $M$ should be Sulphur $(S)$
$∴$ Equvalent weight of $S = \frac{32}{4} = 8$

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A solid element burns completely in oxygen to give gaseous oxide without any change in volume. If the vapour density of oxide is 32, the equivalent weight of element would be?

The correct option is A 82M+O2→2MOMolecular weight of gaseous oxide =2× vapour density =2×32=64∴ So molar mass of M=64−2×16=32So M should be Sulphur (S)∴ Equvalent weight of S=324=8

A solid element burns completely in oxygen to give gaseous oxide without any change in volume. If the vapour density of oxide is $32$ , the equivalent weight of element would be?

The correct option is A $8$
$2 M + O_{2} \to 2 M O$
Molecular weight of gaseous oxide $= 2 \times$ vapour density
$= 2 \times 32 = 64$
$∴$ So molar mass of $M = 64 - 2 \times 16 = 32$
So $M$ should be Sulphur $(S)$
$∴$ Equvalent weight of $S = \frac{32}{4} = 8$