Equal masses of H2 , O2 and methane have been taken in a container of volume V at temperature 27^°C in identical conditions. The ratio of the volumes of gases H2 : O2 : methane would be [AIPMT-2014] (1) 8 : 16 : 1 (2) 16 : 8 : 1 (3) 16 : 1 : 2 (4) 8 : 1 : 2

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Solution

Let us consider that the hydrogen , oxygen and methane are present in 100 grams proportion.
Since the volume of gas is directly proportional to number of moles of gas
The molar mass of hydrogen oxygen and methane are 2 gm/mol , 32 gm/mol and 16 g,/mol respectively.
Step 1 - Calculation of number of moles
$N u m b e r o f m o l e s o f h y d r o g e n = \frac{G i v e n m a s s i n g r a m s}{M o l e c u l a r m a s s} = \frac{100}{2} = 50 m o l e s$
$N u m b e r o f m o l e s o f o x y g e n = \frac{G i v e n m a s s}{M o l e c u l a r m a s s} = \frac{100}{32} = \frac{1}{16}$
$N u m b e r o f m o l e s o f m e t h a n e = \frac{G i v e n m a s s}{M o l e c u l a r m a s s} = \frac{100}{16} = \frac{1}{8}$
$T h e r a t i o o f v o l u m e o f h y d r o g e n o x y g e n a n d m e t h a n e w i l l b e h e n c e$
$\frac{100}{2} : \frac{100}{32} : \frac{100}{16} = \frac{1}{1} : \frac{1}{16} : \frac{1}{8} = 16 : 1 : 2$

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