Question

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

Answer 1

No of nuetrons in one atom $=(30-15)=15$

0.18g ${}_{15}^{30}P=0.18/30=0.006mole$

no of ${}_{15}^{30}P$ in 0.006 mole=$0.006X6.02X10^{23}$

No of electrons in 0.006 mole of ${}_{15}^{30}P$ =15x

$0.006X6.02X10^{23}$$=5.418\times 10^{22}$ and no of nuetrons $=5.418\times 10^{22}$