Question

The volume occupied by 1.6gm of oxygen at NTP is

Answer 1

Answer:

PLEASE GIVE ME A BRANLIAST ANSWER

As the volume occupied by 1mole of any gas under NTP is 22.4 L

and 1 mole of O2=32 g

so 16g=0.5 mole

hence 0.5mole of O2 will occupy 22.4/2 L of volume

i.e., 11.2 L.

Answer 2

The volume of oxygen occupied by oxygen at NTP is 1.12litre.

Given,

Amount of oxygen= 1.6 grams at NTP.

To Find,

The volume of oxygen occupied by oxygen at NTP.

Solution,

We can solve this type of question by using these ways.

Amount of oxygen= 1.6 gram

The molecular mass of oxygen = 32 gram

Now, 32-gram oxygen occupied a 22.4-litre volume at NTP

So 1.6-gram of oxygen at NTP would occupy a volume

$= \frac{22.4}{32} \times 1.6$

$= 1.12d {m}^{3}$

Hence, The volume of oxygen occupied by oxygen at NTP is 1.12litre.