Question

Calculate the following, present in 16 gm of ${O_2}$
A. Numbers of moles
B. Number of molecules
C. Number of gram atoms
D. Number of atoms
E. Volume occupied by it at S.T.P.
F. Number of protons

Answer 1

Hint: Molecular weight is the average weight of the compound compared to 1/12th mass of carbon, Atomic number is the number of protons present in the nucleus, Avogadro number is the number of atoms or molecules present in 1 mole of a substance.

 Complete step by step answer:
-Number of moles:
We know that the molecular weight of oxygen is 32.
Thus the number of moles present in 16 g of oxygen will be = $\dfrac{{{\text{Given weight}}}}{{{\text{Molecular weight}}}} = \dfrac{{16}}{{32}} = 0.5$

-Number of molecules:
From above we know that number of molecules present in oxygen will be = ${\text{Number of moles }} \times {\text{ }}{N_A} = 0.5 \times {N_A}$

-Number of gram atoms:
Number of gram atoms = \[0.5 \times 2{N_A} = {N_A}\]​ (because one molecule of oxygen contains 2 atoms of oxygen).

-Number of atoms:
Number of atoms is equal to number of moles, therefore from part (A) number of atoms = 0.5

-Volume occupied by it at S.T.P.
We know that 1 mole of a gas occupies 22.4L at S.T.P.,
Therefore 0.5 mole of gas will occupy: ${\text{Number of moles }} \times {\text{ Volume at S}}{\text{.T}}{\text{.P = }}0.5 \times 22.4 = 11.2{\text{ L}}$
Thus given sample will occupy 11.2L at STP

-Number of protons:
We know that number of protons = Number of gram atoms × 8 (because one oxygen atom contains 8 protons), Therefore from (C) we know that number of gram atoms = ${N_A}$
Therefore, number of protons = $8 \times {N_A}$

Note: 273 K and one atmosphere or 760 mm of Hg or 76 cm of Hg pressure are known as the standard conditions of temperature and pressure (STP) or normal condition of temperature and pressure (NTP).
The mass of one mole atom of any element is exactly equal to the atomic mass in grams of that element. And it depends on the nature of particles