If $V$ ml of the vapours of substance at NTP weighs $W$ g. Then, molecular weight of substance is :

\frac{W}{V} \times 22400

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\frac{V}{W} \times 22.4

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(W - V) \times 22400

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\frac{W \times 1}{V \times 22400}

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Solution

The correct option is A $\frac{W}{V} \times 22400$
Number of mole $= \frac{g i v e n v o l u m e}{m o l a r v o l u m e} = \frac{V}{22400}$
Mole $= \frac{m a s s}{m o l a r m a s s} = \frac{W}{M o l . w t .}$
$\frac{V}{22400} = \frac{W}{M o l . w t .}$

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If V ml of the vapours of substance at NTP weighs W g. Then, molecular weight of substance is :

The correct option is A WV×22400Number of mole =given volumemolar volume=V22400Mole =massmolar mass=WMol.wt.V22400=WMol.wt.

If $V$ ml of the vapours of substance at NTP weighs $W$ g. Then, molecular weight of substance is :

The correct option is A $\frac{W}{V} \times 22400$
Number of mole $= \frac{g i v e n v o l u m e}{m o l a r v o l u m e} = \frac{V}{22400}$
Mole $= \frac{m a s s}{m o l a r m a s s} = \frac{W}{M o l . w t .}$
$\frac{V}{22400} = \frac{W}{M o l . w t .}$