Question

0.44 gram of gas occupies 224 ml at STP its vapour density is​

Answer 1

Answer:

Molecular mass=2*vapour density 

=2*11.2

=22.4 g

At NTP,

volume occupied by 22.4 g of gas= 22.4 l

volume of 11.2g gas = 22.4/22.4*11.2

= 11.2 l

Answer 2

Answer:

The vapor density of the given gas under the circumstances would be 11.2.

Explanation:

From our knowledge of molecular chemistry, we understand that $Molecular\;mass = 2 \times vapour\;density$

We are also given that the mass of the gas is 0.44 g, and the volume of the gas is 224 mL.

Thus, we can solve this as $Molecular\;mass = 2 \times 11.2\\Molecular\;mass = 22.4 g$

At normal temperature and pressure, the volume occupied by 22.4 grams of gas is 22.4 mL.
Thus, the vapor density would be 11.2

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