Question

Number of electrons present in $ 3.6\, mg $ of $ NH^+_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 $NH^{+}_{4}(W)=3.6\,mg$
$=3.6\times 10^{-3}\,g$
Molar mass of $NH^{+}_{4}(M)=18\,g$
$\because \frac{w}{M}=\frac{N}{N_{A}}$, where $N$ =Number of $NH_{4}^{+}$ ion
$\therefore N=\frac{W\times N_{A}}{M}$
$\frac{3.6\times10^{-3}\times6.02\times10^{23}}{18}$
$=1.20\times10^{20}$
$\therefore $ No. of electron in one $NH_{4}^{+}=10\left(7+4+1\right)$
$\therefore $ Total no. of electrons $=10\times1.2\times10^{20}$
$=1.2\times10^{21}$