Question

Amount of ${\text{8}}$ g of barium carbonate ${\text{BaC}}{{\text{O}}_{\text{3}}}$ is needed to kill a rat. How many molecules is this?

Answer 1

We will use the Avogadro number to determine the number of molecules in the given gram. For this first, we have to calculate the moles of barium carbonate in the given gram. For this, we will use the mole formula. Then by using the Avogadro number we will determine the numbers of molecules.

Complete step by step answer:
We will use the mole formula to determine the number of moles of barium carbonate as follows:
${\text{mole}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{{\text{molar}}\,{\text{mass}}}}$
Molar mass of barium carbonate is $197.34$ g/mol.
On substituting ${\text{8}}$ grams for mass of barium carbonate and $197.34$ for molar mass of barium carbonate,
${\text{mole}}\,{\text{ = }}\,\dfrac{8}{{197.34}}$
${\text{mole}}\,{\text{ = }}\,0.041$
So, the moles of the barium carbonate is $\,0.041$.
According to the Avogadro number,
One mole of any substance =$\,6.02 \times {10^{23}}$ molecules
So,
One mole =$\,6.02 \times {10^{23}}$ molecules of barium carbonate
So, $\,0.041$ moles of barium carbonate will have,
$\,0.041$ mole of barium carbonate =$\,6.02 \times {10^{23}}\, \times \,\,0.041$ molecules of barium carbonate
= $2.44 \times {10^{22}}$ molecules of barium carbonate
So, $2.44 \times {10^{22}}$ barium carbonate molecules are there in $8$ g of barium carbonate.

Therefore, $2.44 \times {10^{22}}$ molecules is the correct answer.

Note: The number of atoms present in $12\,{\text{g}}$ of carbon-12 is known as one mole. In case of monoatomic, one mole of substance contains Avogadro's number of atoms. The subscript after each atom represents the numbers of that atom. The superscript represents the charge of an ion not the number of that ion. We can also determine the total number of atoms here, for this we have to multiply the Avogadro with the number of atoms present in a molecule.