Question

How many protons, electrons and neutrons are present in 0.18g $${}_{15}^{30}P$$?

Answer 1

Hint: 30g is the atomic weight of phosphorus for Avogadro number of atoms. Try to find out the number of moles present in 0.18g of Phosphorus. Then calculate the number of atoms present in 0.18g Phosphorous.

Complete Answer:
Phosphorous, P is an element in the periodic table having the atomic number, $Z=15$ and atomic mass number, $A=30$.
- We know that in an atom, number of electrons = number of protons =$Z=15$
- Number of neutrons in an atom is the difference between atomic mass number and atomic number i.e. $(A-Z)=30-15=15$
- 1 mole of P contains Avogadro’s number of atoms, $N_A$ $$(6.023\times {10}^{23})$$ which is equal to 30g.
- So now, we need to calculate the number of moles of Phosphorus are present in 0.18g.
$$1\text{ mole}=30g$$
$${}^{\prime }{x}^{\prime }\text{moles}=0.18g$$
- By cross-multiplication we obtain,
$$x=\frac{0.18}{30}=6\times {10}^{-3}\text{moles}$$
- Therefore, 0.18g of P contains $6\times {10}^{-3}moles$.
     $$1\text{mole}=N_A\text{atoms}=6.023\times {10}^{23}\text{atoms}$$
     $6\times {10}^{-3}\text{moles}=6\times {10}^{-3}\times N_A\text{atoms}=6\times {10}^{-3}\times 6.023\times {10}^{23}\text{atoms}$
- Thus, 0.18g of P contains $3.614\times {10}^{21}$atoms.
- Now that we have found out the number of atoms present in 0.18g P. Let’s calculate the number of protons.
Number of protons $=15\times 3.614\times {10}^{21}=5.421\times {10}^{22}$
Similarly, number of electrons $=5.421\times {10}^{22}$
$\begin{array}{rl}& \text{Number of neutrons}=(A-Z)\times 3.614\times {10}^{21}\\ & =(30-15)\times 3.614\times {10}^{21}\\ & =15\times 3.614\times {10}^{21}\\ & =5.421\times {10}^{22}\end{array}$
Therefore, the total number of protons, electrons and neutrons is the same which is equal to $5.421\times {10}^{22}$.

Note:
Remember atomic mass number (Z) gives the mass of 1 mole of atoms of any element present in the periodic table. Don’t get confused with the term “mole” and “molecule”. One mole contains Avogadro’s number of atoms/molecules i.e. $6.023\times {10}^{23}$.