Question

How many phosphorus atoms are present in $0.15$ mole of ${P_4}$ molecules?
a.$1.67 \times {10^{22}}$
b.$3.6 \times {10^{23}}$
c.$0.60$
d.$6.6 \times {10^{23}}$

Answer 1

Hint: Phosphorus is the element whose atomic number is $15$ which indicates it has $15$ protons and $15$ electrons. It is represented by the symbol $P$. $1$ mole of ${P_4}$ has $6.023 \times {10^{23}}$ number of atoms and $1$ molecule of ${P_4}$ have $4$ atoms of $P$.

Complete answer:
This law is also termed as Avogadro’s hypothesis. According to this law, the total quantity of atoms or molecules of the gas is directly proportional to the volume occupied by the gas at constant temperature and pressure. It is firmly connected to the ideal gas equation as it associates temperature, pressure, volume and the amount of substance for a given gas. We get this relation from the kinetic theory of gases under the assumption of the conditions of an ideal gas.
Avogadro’s law states that one mole of a matter or element is equivalent to $6.023 \times {10^{23}}$ number of units of matter like atoms or molecules or ions. This specific value i.e. $6.023 \times {10^{23}}$ is termed as Avogadro’s number or Avogadro’s constant.
In this question, the number of atoms present in the ${P_4}$ molecule is four. Thus,
$1$ mole of ${P_4}$ have :
$ \Rightarrow number\;of\;atoms\,\;in\,\;{P_4} \times \;Avogadro\;number$
$ \Rightarrow 4 \times 6.023 \times {10^{23}}\,P\;atoms$
Thus, $0.15$ mole of ${P_4}$ have:
$ \Rightarrow 4 \times 6.023 \times {10^{23}} \times 0.15$
$ \Rightarrow 3.61 \times {10^{23\,}}\;P\,atoms$

Thus, the correct answer is option $\left( b \right)$.

Note:
Avogadro constant’s value was taken because to equalize the mass of one mole of compound which is in grams and the average mass of a molecule of the compound which is in daltons. So, it is the proportionality component that connects the molar mass of a matter to the average mass of a molecule and is also the approximate value of nucleons present in one gram of common matter.