Question

How many molecules are present in${\text{1296}}\,{\text{g}}$ of dinitrogen pentoxide ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$?

Answer 1

The mole concept is used to determine the amount of the substance which is obtained by taking the ratio of the mass of the substance to the molecular mass of the substance.
One mole of the substance contains the Avogadro’s numbers of the particles that are atoms, ions, or molecules of the substance.Avogadro’s number is represented as ${{\text{N}}_{\text{A}}}$. The value of Avogadro’s number is ${\text{6}}{{.023 \times 1}}{{\text{0}}^{{\text{23}}}}\,{\text{molecules}}$.
The molecular weight of the dinitrogen pentoxide is ${\text{108}}{\text{.01}}\,{\text{g/mol}}$.

Formula used: The mole of the substance contains Avogadro’s numbers of molecules.
${\text{1mole}}\,{\text{ = }}\,{{\text{N}}_{\text{A}}}{\text{ = 6}}{{.023 \times 1}}{{\text{0}}^{{\text{23}}}}\,{\text{molecules}}$

Complete step-by-step solution: Here, the molecular weight of the ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ given is ${\text{108}}{\text{.01}}\,{\text{g/mol}}$ and the weight of the ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ is 1296 grams.
We have to determine the first moles of the ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ in 1296 gram as follows:
${\text{moles = }}\dfrac{{{\text{weight}}}}{{{\text{Molecular}}\,{\text{weight}}}}$
Substitute, ${\text{108}}{\text{.01}}\,{\text{g/mol}}$ for molecular weight and 1296 gram for weight.
${\text{moles = }}\dfrac{{1296\,{\text{g}}}}{{{\text{108}}{\text{.01}}\,{\text{g/mol}}}}$
$\Rightarrow {\text{moles = 11}}{\text{.9989}}\,{\text{mol}}$
Thus, the moles of the ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ in 1296 gram are ${\text{11}}{\text{.9989}}\,{\text{mol}}$.
Now, determine the molecules of the ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$ in the 1296 gram as follows:As we know that one mole of the substance contains Avogadro’s numbers of molecules.
${\text{1mol}}\,{\text{ = }}\,{{\text{N}}_{\text{A}}}{\text{ = 6}}{{.023 \times 1}}{{\text{0}}^{{\text{23}}}}\,{\text{molecules}}$
$\Rightarrow {\text{11}}{\text{.9989}}\,{\text{mol = }}\dfrac{{{\text{11}}{\text{.9989}}\,{{mol \times 6}}{{.023 \times 1}}{{\text{0}}^{{\text{23}}}}\,{\text{molecules}}}}{{{\text{1mol}}\,}}$
$\therefore {\text{11}}{\text{.9989}}\,{{mol = 7}}{{.2269 \times 1}}{{\text{0}}^{{\text{24}}}}\,{\text{molecules}}$

Thus, the molecules of ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$in 1296 grams are ${\text{7}}{{.2269 \times 1}}{{\text{0}}^{{\text{24}}}}$.

Note: Molecular weight is the average mass of the molecules and it is deferrable from molecule to molecule. In the case of the atom, it is called atomic mass while in the case of the molecular mass it is calculated by taking the sum of the atomic masses of all the atoms in the molecule.Here, we have to use the molecular weight of the ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}$. The SI unit of the molecular weight is gram per mole. Its unit itself indicates the ratio of the mass of the substance to the molar molecular mass of the substance.