$0.5 m o l e$ of a gas (mol. wt. = 20) occupies $11.2 L$ at $S T P$ . The volume occupied by $0.25 m o l e$ of a lighter gas (mol. wt. = 2) at $S T P$ will be:

11.2

L

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5.6

L

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2.8

L

No worries! We‘ve got your back. Try BYJU‘S free classes today!

22.4

L

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 5.6 $L$
At $S T P$ , the volume occupied is directly proportional to the number of moles of a gas.

$1$ mole of any gas equals to $22.4 L$ of gas at STP.
$\frac{11.2 L o f h e a v i e r g a s}{0.5 m o l e o f h e a v i e r g a s} = \frac{V o l u m e o f l i g h t e r g a s}{0.25 m o l e o f l i g h t e r g a s}$
$∴$ Volume of lighter gas $= \frac{11.2}{0.5} \times 0.25 = 5.6 L$

Hence, the correct answer is option $B$ .

Suggest Corrections

Similar questions

Q. A gas has a vapour density of

11.2

. The volume occupied by

1 g

of the gas at STP is:

Q. The vapour density of a gas is 11.2. The volume occupied by 10 g of the gas at STP is

Q. One mole of a light gas (mol. wt.

=

32) occupies a volume of

20 L

under certain conditions. Under the same conditions two moles of a heavy gas (mol. wt.

=

64) occupies a volume of:

The vapour density of a gas is 11.2. The volume occupied by 11.2 g of the gas at ATP will be

If one mole of a gas (mol.wt-40) occupies a volume of 20litres, under the same conditions of temperature and pressure the volume occupied by 2 moles of gas B (mol.wt=80) is

Join BYJU'S Learning Program

0.5mole of a gas (mol. wt. = 20) occupies 11.2L at STP. The volume occupied by 0.25mole of a lighter gas (mol. wt. = 2) at STP will be:

$0.5 m o l e$ of a gas (mol. wt. = 20) occupies $11.2 L$ at $S T P$ . The volume occupied by $0.25 m o l e$ of a lighter gas (mol. wt. = 2) at $S T P$ will be: