Question

An oxide of N has vapour density 46. Find the total number of electrons in its 92 g.(N${}_A$ = Avogadro's number)

• A. 46 N${}_A$
• B. 38 N${}_A$
• C. 54 N${}_A$
• D. 30 N${}_A$

Answer 1

The correct option is A 46 N${}_A$

Vapour density = 46
Molecular weight = 92
So, compound is $N_2{O}_4$.
Total number of e${}^{-}$ = (7 $\times$ 2) + (8 $\times$ 4) = 46
So, 92 gm of compound means one mole will have 46$N_A$ electrons.