The number of atoms present in 62 g of phosphorous if its gram atomic mass is 31 g is p $\times {10}^{24}$. What is the value of p?

1. 1.204

The correct option is A 1.204
Given, mass of phosphorus = 62 g
The gram atomic mass (GAM) of phosphorus = 31 g

$\therefore$ Number of moles of phosphorus
= $\frac{\text{Given mass in grams}}{\text{G.A.M}}$

Number of moles = $\frac{62}{31}$ = 2 moles

Each mole has Avogadro's number of particles
( $N_A$ = $6.023\times {10}^{23}$)

$\therefore$ The number of atoms present in 62 grams of phosphorous = $2\times N_A$
= $2\times 6.023\times {10}^{23}$
= $12.04\times {10}^{23}$
= $1.204\times {10}^{24}$

$\therefore$ The value of p (in p $\times {10}^{24}$) = 1.204