The vapour density of a gas is given by :

V . D =

\frac{M o l a r m a s s}{2}

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V . D =

\frac{M a s s o f N m o l e c u l e s o f g a s}{M a s s o f N m o l e c u l e s o f H_{2}}

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V . D =

\frac{M a s s o f 1 m o l e c u l e o f g a s}{M a s s o f 1 m o l e c u l e o f H_{2}}

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None of the above

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Solution

The correct options are
B $V . D =$ $\frac{M o l a r m a s s}{2}$
C $V . D =$ $\frac{M a s s o f N m o l e c u l e s o f g a s}{M a s s o f N m o l e c u l e s o f H_{2}}$
D $V . D =$ $\frac{M a s s o f 1 m o l e c u l e o f g a s}{M a s s o f 1 m o l e c u l e o f H_{2}}$
Vapour Density of a gas $= \frac{D e n s i t y o f t h e g a s}{D e n s i t y o f h y d r o g e n}$
$= \frac{M a s s o f c e r t a i n v o l u m e o f t h e g a s a t S T P}{M a s s o f t h e s a m e v o l u m e o f H}$
$= \frac{M a s s o f n m o l e c u l e s o f t h e g a s}{M a s s o f n m o l e c u l e s o f H_{2}}$
$= \frac{M a s s o f 1 m o l e c u l e o f t h e g a s}{M a s s o f 1 m o l e c u l e (o r 2 a t o m s o f H) o f h y d r o g e n g a s}$
$= \frac{M o l e c u l a r m a s s}{2}$
Hence the correct options are A, B and C.

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(a) Oxygen is a diatomic molecule.
(b) 2 × V.D. = Molecular mass.
(c) One mole of any gas contains the same number of molecules.

C H_{4}

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