Question

number of P atoms present in 0.5g molecules of P4 is​

Answer 1

Explanation:

no. of atoms= n×NA×atomicity

Answer 2

Answer : The number of phosphorous atoms are, $9.707\times 10^{21}$

Explanation : Given,

Mass of $P_4$ = 0.5 g

Molar mass of $P_4$ = 124 g/mole

First we have to calculate the moles of $P_4$

$\text{Moles of }P_4=\frac{\text{Mass of }P_4}{\text{Molar mass of }P_4}=\frac{0.5g}{124g/mole}=0.00403mole$

As we know that $P_4$ contains 4 number of phosphorous atoms.

Now we have to calculate the number of atoms of phosphorous.

As, 1 mole of $P_4$ contains $4\times 6.022\times 10^{23}$ number of phosphorous atoms

As, 0.00403 mole of $P_4$ contains $0.00403\times 4\times 6.022\times 10^{23}=9.707\times 10^{21}$ number of phosphorous atoms

Therefore, the number of phosphorous atoms are, $9.707\times 10^{21}$