Question

The number of atoms present in one mole of an element is equal to Avogadro number. Which of the following elements contains the greatest number of atoms?
A) $4{\text{ g He}}$
B) $46{\text{ g Na}}$
C) $0.40{\text{ g Ca}}$
D) $12{\text{ g He}}$

Answer 1

We know that the amount of substance having exactly the same number of atoms as are present in $12{\text{ g}}$ of ${{\text{C}}^{{\text{12}}}}$ is known as mole. Moles is the ratio of the mass of substance in g to the molar mass of the substance in ${\text{g/mol}}$. ${\text{1 mol}}$ of any substance contains $6.022 \times {10^{23}}$ atoms. $6.022 \times {10^{23}}$ is Avogadro's number.

Formula Used: ${\text{Number of moles}}\left( {{\text{mol}}} \right){\text{ = }}\dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}$

Complete step-by-step answer:
Calculate the number of moles of helium in $4{\text{ g He}}$ using the formula for the number of moles.
We know that the molar mass of helium is $2{\text{ g/mol}}$. Thus,
${\text{Number of moles of helium}} = \dfrac{{{\text{4 g}}}}{{{\text{2 g/mol}}}} = 2{\text{ mol}}$
Thus, the number of moles of helium in $4{\text{ g He}}$ is $2{\text{ mol}}$.
We know that ${\text{1 mol}}$ of any substance contains $6.022 \times {10^{23}}$ atoms. Thus,
${\text{Number of helium atoms}} = 2{\text{ mol}} \times \dfrac{{6.022 \times {{10}^{23}}{\text{ atoms}}}}{{1{\text{ mol}}}} = 12.044 \times {10^{23}}{\text{ atoms}}$
Thus, $4{\text{ g He}}$ contains $12.044 \times {10^{23}}{\text{ atoms}}$ of helium.
Calculate the number of moles of sodium in $46{\text{ g Na}}$ using the formula for the number of moles.
We know that the molar mass of sodium is $23{\text{ g/mol}}$. Thus,
${\text{Number of moles of sodium}} = \dfrac{{{\text{46 g}}}}{{{\text{23 g/mol}}}} = 2{\text{ mol}}$
Thus, the number of moles of sodium in $46{\text{ g Na}}$ is $2{\text{ mol}}$.
We know that ${\text{1 mol}}$ of any substance contains $6.022 \times {10^{23}}$ atoms. Thus,
${\text{Number of sodium atoms}} = 2{\text{ mol}} \times \dfrac{{6.022 \times {{10}^{23}}{\text{ atoms}}}}{{1{\text{ mol}}}} = 12.044 \times {10^{23}}{\text{ atoms}}$
Thus, $46{\text{ g Na}}$ contains $12.044 \times {10^{23}}{\text{ atoms}}$ of sodium.
Calculate the number of moles of calcium in $0.40{\text{ g Ca}}$ using the formula for the number of moles.
We know that the molar mass of calcium is $40{\text{ g/mol}}$. Thus,
${\text{Number of moles of calcium}} = \dfrac{{{\text{0}}{\text{.40 g}}}}{{{\text{40 g/mol}}}} = 0.01{\text{ mol}}$
Thus, the number of moles of calcium in $0.40{\text{ g Ca}}$ is $0.01{\text{ mol}}$.
We know that ${\text{1 mol}}$ of any substance contains $6.022 \times {10^{23}}$ atoms. Thus,
${\text{Number of calcium atoms}} = 0.01{\text{ mol}} \times \dfrac{{6.022 \times {{10}^{23}}{\text{ atoms}}}}{{1{\text{ mol}}}} = 0.06022 \times {10^{23}}{\text{ atoms}}$
Thus, $0.40{\text{ g Ca}}$ contains $0.06022 \times {10^{23}}{\text{ atoms}}$ of calcium.
Calculate the number of moles of helium in $12{\text{ g He}}$ using the formula for the number of moles.
We know that the molar mass of helium is $2{\text{ g/mol}}$. Thus,
${\text{Number of moles of helium}} = \dfrac{{12{\text{ g}}}}{{{\text{2 g/mol}}}} = 6{\text{ mol}}$
Thus, the number of moles of helium in $12{\text{ g He}}$ is $6{\text{ mol}}$.
We know that ${\text{1 mol}}$ of any substance contains $6.022 \times {10^{23}}$ atoms. Thus,
${\text{Number of helium atoms}} = 6{\text{ mol}} \times \dfrac{{6.022 \times {{10}^{23}}{\text{ atoms}}}}{{1{\text{ mol}}}} = 36.132 \times {10^{23}}{\text{ atoms}}$
Thus, $12{\text{ g He}}$ contains $36.132 \times {10^{23}}{\text{ atoms}}$ of helium.
Thus, the element containing the greatest number of atoms is $12{\text{ g He}}$.
Thus, the correct option is (D) $12{\text{ g He}}$.

Note: The number of atoms of a compound is Avogadro’s number for 1 mole of compound. The number $6.022 \times {10^{23}}$ is known as Avogadro’s number. The Avogadro’s number gives the number of atoms, ions or molecules present in one mole of any substance.