Question

Which is/are correct statements about $1.7$ g of ${NH}_3$?

• A. It contain $0.3$ mole $H$ - atom
• B. It contain $2.408\times {10}^{23}$ atoms
• C. Mass % of $H$ is $17.65$%
• D. It contains $0.3$ mol $N$-atom

Answer 1

Answer: (A), (B), (C)

Mass % of $H$ is $17.65$%
It contain $0.3$ mole $H$ - atom
It contain $2.408\times {10}^{23}$ atoms

Moles of ${NH}_3$ $=0.1$. Then, moles of $H$ atom $=0.3$

Total atoms $=4\times 0.1\times 6.02\times {10}^{23}=2.408\times {10}^{23}$

% $H$ = $\frac{0.3mole\times 1g/mole}{1.7g}\times 100=$ $17.65$ %