Question

Number of electrons present in $3.6mg$ of $N{H}_4^{+}$ are

• A. $1.20\times 10^{21}$
• B. $1.20\times 10^{20}$
• C. $1.20\times 10^{22}$
• D. $2\times 10^{-3}$

Answer 1

Answer: (A)

Given, Mass of $N{H}_4^{+}(W)=3.6mg$
$=3.6\times 10^{-3}g$
Molar mass of $N{H}_4^{+}(M)=18g$
$\because \frac{w}{M}=\frac{N}{N_A}$, where $N$ =Number of $N{H}_4^{+}$ ion
$\therefore N=\frac{W\times N_A}{M}$
$\frac{3.6\times 10^{-3}\times 6.02\times 10^{23}}{18}$
$=1.20\times 10^{20}$
$\therefore$ No. of electron in one $N{H}_4^{+}=10\left(7+4+1\right)$
$\therefore$ Total no. of electrons $=10\times 1.2\times 10^{20}$
$=1.2\times 10^{21}$