Question

An oxide of $N$ has vapour density $46$. Find the total number of electrons in its $92g$.

• A. $46\times 6.023\times {10}^{23}$ electrons
• B. $46$ electrons
• C. $92\times 6.023\times {10}^{23}$ electrons
• D. $6.023\times {10}^{23}$ electrons

Answer 1

The correct option is A $46\times 6.023\times {10}^{23}$ electrons

Let oxide of $N$ be $N_2{O}_m$
$\because$ Vapour density of $N_2{O}_m=46$ = $\frac{molecularmassofgas}{molecularmassofhydrogen}$
$\therefore$ Molar mass of $N_2{O}_m=2\times 46=92g$ $mo{l}^{-1}$
$\because 2\times 14+m\times 16=92$ or $m=4$
$\because$Oxide is $N_2{O}_4$
Now given $92g$ $N_2{O}_4,i.e.,N$ molecules of $N_2{O}_4$
$\because 1$ molecule of $N_2{O}_4$ has $46$ electrons
$\therefore N$ molecules of $N_2{O}_4=46\times N$ electrons
$=46\times 6.023\times {10}^{23}$ electrons