Question

Number of atoms of oxygen present in 11.2l of ozone at ntp

Answer 1

Answer:

One mole of ozone has 6.023×10^23 molecules.

One mole of Ozone has volume 22.4 liter.

So 11.2 liter of ozone is 0.5 mole.

So 0.5 moles of Ozone has 0.5× 6.023×10^23 molecules.

One molecule of ozone has three oxygen atoms.

So total atoms are- 3×0.5×6.023×10^23=9.03×10^23 numbers.

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Hope it helps.

Answer 2

Number of atoms of Oxygen present in 11.2L of Ozone at NTP are 9.03* 10²³ atoms.

Given:-

Volume of Ozone at NTP = 11.2L

To Find:-

Number of atoms of Oxygen present in 11.2L of Ozone at NTP.

Solution:-

We can simply find out the Number of atoms of Oxygen present in 11.2L of Ozone at NTP by following these simple steps.

As,

Volume of Ozone at NTP (v1) = 11.2L

Actual Volume (v) = 22.4L

So, before going to the Number of atoms of Oxygen present in 11.2L of Ozone at NTP. Forat we will calculate the number of moles of Oxygen.

According to the formula,

number of moles (n) = volume (v1) / 22.4

$n = \frac{v1}{v}$

$n = \frac{11.2}{22.4}$

$n = \frac{1}{2}$

$n = 0.5$

Now,

Number of atoms in 1 mol O2 = 6.02* 10²³

Number of atoms in 0.5 mol O2 = 6.02* 10²³ *0.5

Number of atoms in 0.5 mol Ozone = 6.02* 10²³ *0.5*3

N = 18.06*0.5*10²³

N = 9.03* 10²³ atoms

Hence, Number of atoms of Oxygen present in 11.2L of Ozone at NTP are 9.03* 10²³ atoms.

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