Question

How many molecules and atoms of oxygen are present in 5.6L of oxygen at NTP?​

Answer 1

5.6L of O2 means we have 0.25 moles of O2.  

As, 1 mole has 6.023*10^23 molecules,

0.25 moles of O2 will have 0.25*6.023*10^23 molecules=1.50575*10^23 molecules

and as 1 molecule of O2 has 2 atoms, so, 1.50575*10^23 molecules will have 2*1.50575*10^23 atoms=3.0115*10^23 atoms of O.

Explanation:

At NTP , one mole of a gas occupies a volume of 22.4 L . This particular volume is known as molar volume .

Also we know that one mole of a gas contains avogadro number of molecules and it contains atoms equal to the product of number of molecules and atomicity .

The atomicity of oxygen is 2 , so 1 mole of oxygen contains atoms equal to 2 times the number of molecules

Therefore we can say that 22.4 L of Oxygen contains = 6.023 * 10^(23) molecules

1 L of Oxygen contains = 6.023 * 10^(23)/22.4 molecules

5.6 L of Oxygen contains = 6.023 * 10^(23) * 5.6/22.4 molecules

5.6 L of Oxygen contains = 6.023 * 10^(23)/4 molecules

5.6 L of Oxygen contains = 1.50575 * 10^(23) molecules

Therefore , 5.6 L of Oxygen contains = 1.50575 * 10^(23) * 2 atoms

5.6 L of Oxygen contains = 3.0115 * 10^(23) atoms

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Answer 2

Explanation:

no.of atoms,

Molar volume of Oxygen =11.2L

Hence, 11.2L contains 6.022×10²³  atoms.

Hence for 5.6L, no. of atoms =6.022×10²³ x $\frac{5.6}{11.2}$

No. of atoms =3.01×10²³ atoms

no.of molecules,

Na molecules ====> 22.4 L

x molecules ======> 5.6 L

x = (Na x 5.6) / 22.4

x = Na /4 molecules

where Na ===> Avogadro number --> 6.022 x 10²³

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