22.4 litres of a gas at S.T.P. weigh 16 g. Identify the gas.

C H_{4}

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O_{2}

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N_{2}

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All of these

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Solution

The correct option is A $C H_{4}$
Any gas can be identified from its molecular weight.
$V o l u m e o c c u p i e d b y t h e g a s a t t h e S . T . P . = 22.4 l i t r e s$
$M a s s o f t h e g a s = 1.6 g .$
$M o l e c u l a r w e i g h t = ?$
Terms involved in the problems are weight and volume of gas at S.T.P.
The standard relation connecting the above two terms is:
$G M W o f a n y g a s o c c u p i e s - ---- \to 22.4 l i t r e s a t S . T . P .$
$2.24 l i t r e s o f t h e g i v e n g a s a t S . T . P . w e i g h - -- \to 1.6 g$ .
$22.4 l i t r e s o f a n y g a s a t S . T . P . w e i g h - -- \to G M W$
$\Rightarrow G M W = \frac{1.6 \times 22.4}{2.24} = 16 g$
$\Rightarrow M o l e c u l a r w e i g h t = 16.$
Therefore, the molecular weight of the gas is 16 and the gas is methane $(C H_{4})$

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Similar questions

Q. 22.4 litres of a gas at STP weighs 16 g. Identify the gas.

22.4

litres of a gas weighs

70

g at S.T.P. Calculate the weight (gm) of the gas if it occupies a volume of

20

litres at

27^{o}

C and

700

mm Hg of pressure.

22.4 litres of a gas weighs 70 g at S.T.P. Calculate the weight of the gas if it occupies a volume of 20 litres at $27^{\circ} C$ and 700 mm Hg of pressure.

C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O

Initially,

16 g

C H_{4}

is combusted in

22.4 L

O_{2}

at STP. Identify the correct statement.

^{'} x^{'}

moles of

N_{2}

gas at S.T.P. conditions occupy a volume of 10 litres, then the volume of

^{'} 2 x^{'}

moles of

C H_{4}

273^{o} C

and 1.5 atm is:

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