Question

How many atoms are in ${\text{1}}{\text{.45}}\,{\text{g}}$ of Ag?

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 ${\text{6}}{.023 \times 1}{{\text{0}}^{{\text{23}}}}\,{\text{molecules}}$.
Here, we have to find out the first moles of the silver and then calculate atoms of the silver for the given mass of the silver.

Complete solution:
Here, the molecular weight of the silver$\left( {{\text{Ag}}} \right)$ given is $107.87\,{\text{g/mol}}$ and the weight of the silver is ${\text{1}}{\text{.45}}\,{\text{g}}$.
We have to determine the first moles of the silver in ${\text{454grams}}$as follows:
${\text{moles = }}\dfrac{{{\text{weight}}}}{{{\text{Molecular}}\,{\text{weight}}}}$
Here, the atomic weight of the silver is used instead of the molecular weight.
Substitute, $107.87\,{\text{g/mol}}$ for atomic weight and ${\text{1}}{\text{.45}}\,{\text{g}}$for weight.
${\text{moles = }}\dfrac{{{\text{1}}{\text{.45}}\,{\text{g}}}}{{107.87\,{\text{g/mol}}}}$
${\text{moles = 0}}{\text{.01344}}\,{\text{mol}}$
Thus, the moles of the silver in ${\text{1}}{\text{.45}}\,{\text{g}}$are ${\text{0}}{\text{.01344}}\,{\text{mol}}$.
Now, determine the atoms of the silver in ${\text{1}}{\text{.45}}\,{\text{g}}$as follows:
As we know that one mole of the substance contains Avogadro’s numbers of atoms.
${\text{1mol}}\,{\text{ = }}\,{{\text{N}}_{\text{A}}}{\text{ = 6}}{.023 \times 1}{{\text{0}}^{{\text{23}}}}\,{\text{atoms}}$
${\text{0}}{\text{.01344}}\,{\text{mol = }}\dfrac{{{\text{0}}{\text{.01344}}\,{mol \times 6}{.023 \times 1}{{\text{0}}^{{\text{23}}}}\,{\text{atoms}}}}{{{\text{1mol}}\,}}$
${\text{0}}{\text{.01344}}\,{\text{mol = 8}}{.0949 \times 1}{{\text{0}}^{{\text{21}}}}\,{\text{atoms}}$
Thus, the atoms of silver in ${\text{1}}{\text{.45}}\,{\text{g}}$ are ${\text{7}}{.988 \times 1}{{\text{0}}^{{\text{23}}}}$.

Note: Molecular weight is the average mass of the molecules and it differs 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.