Question

Which of the following contain the maximum number of atoms and why.
(A) 4g ${H_2}$
(B) 16g ${O_2}$
(C) 28g of ${N_2}$
(D) 18g of ${H_2}O$

Answer 1

Hint : In chemistry irrespective of type of elements one mole of any of the elements contains the same number of atoms. The actual number of atoms present in 1 mole is described in the term of Avogadro number.
Avogadro number or Avogadro constant is symbolized by ${N_A}$ .
Value of the Avogadro number is always constant $6.022 \times {10^{23}}$ atoms.

Complete Step By Step Answer:
To calculate the number of atoms present in $4g$ of ${H_2}$ , first we calculate the number of moles of hydrogen present in $4g$ .
As we know mole is calculated by ratio of mass of given compound to the molecular mass of compound.
Given mass of element is $4g$ and molecular mass of hydrogen is $2g$
$m = \dfrac{4}{2}$
$m = 2$
Hence $4g$ of hydrogen contains $2$ mole of hydrogen.
We already know that $1mole = 6.022 \times {10^{23}}$ atoms
Hence, $4g$ of hydrogen contains $6.022 \times {10^{23}} \times 2$ of atoms.
$4g$ of hydrogen contains $12.044 \times {10^{23}}$ atoms.
To calculate the number of atoms present in $16g$ of ${O_2}$ , first we calculate the number of moles of oxygen present in $16g$ .
As we know mole is calculated by ratio of mass of given compound to the molecular mass of compound.
Given mass of element is $16g$ and molecular mass of oxygen is $32g$
$m = \dfrac{{16}}{{32}}$
$m = 0.5$
Hence $16g$ of oxygen contains $0.5$ mole of oxygen.
We already know that $1mole = 6.022 \times {10^{23}}$
Hence, $16g$ of oxygen contains $6.022 \times {10^{23}} \times 0.5$ of atoms.
$16g$ of oxygen contains $3.011 \times {10^{23}}$ atoms.
To calculate the number of atoms present in $28g$ of nitrogen, first we calculate the number of moles of nitrogen present in $28g$ .
As we know mole is calculated by ratio of mass of given compound to the molecular mass of compound.
Given mass of element is $28g$ and molecular mass of nitrogen is $28g$
$m = \dfrac{{28}}{{28}}$
$m = 1$
Hence $28g$ of nitrogen contains $1$ moles of nitrogen.
We already know that $1mole = 6.022 \times {10^{23}}$ atoms.
Hence, $28g$ of nitrogen contains $= 6.022 \times {10^{23}}$ of atoms.
To calculate the number of atoms present in $18g$ of water, first we calculate the number of moles of water present in $18g$ .
As we know, mole is calculated by the ratio of mass of a given compound to the molecular mass of a compound.
Given mass of element is $18g$ and molecular mass of water is $18g$
$m = \dfrac{{18}}{{18}}$
$m = 1$
Hence $18g$ of water contains $1$ mole of water.
We already know that $1mole = 6.022 \times {10^{23}}$ atoms.
Hence, $18g$ of water contains $6.022 \times {10^{23}}$ of atoms.
From the above calculation we find that $4g$ of hydrogen contains the maximum number of atoms $12.044 \times {10^{23}}$ atoms.
Hence, option $\left( i \right)$ is correct.

Note :
Remember to convert the weight of elements into moles before calculating the number of atoms in it. One mole of hydrogen contains two atoms of hydrogen, so the molecular mass of hydrogen is considered as $2g$ instead of $1g$ . Consider the atomic mass or molecular mass of any element in grams instead of amu to obtain the mole with no unit.