Question

Calculate the no.of electrons present in 1.4 g of di-nitrogen gas.

Answer 1

Molar mass of $N_2$ = 28 g/mol

]So, no. of moles = 1.4/28 = 0.05 mole

1 mole of any molecule has $6.022\times {10}^{22}$ molecules,

0.05 moles of $N_2$ = $0.05\times$$6.022\times {10}^{22}$

1 molecule of nitrogen has 14 electrons in it.

So,

$0.05\times$$6.022\times {10}^{22}$ molecules = $0.05\times$$6.022\times {10}^{22}\times 14$ electrons

= $4.214\times {10}^{23}$ electrons