Question

Calculate the number of atoms of hydrogen present in $5.6g$ of urea (molar mass of urea $=60gmo{l}^{-1}$).

Answer 1

The molecular formula of urea is $CO{N}_2{H}_2$
Molar mass of urea = $60gmo{l}^{-1}$
Number of moles of urea = $n=\frac{W}{M}$
=$\frac{5.6}{60}=0.09333mol$
Now,
$1$ mol urea contains $2$-gram atoms of hydrogen$\left(H\right)$
$\therefore$ 0.09333mol urea contains=$=2\times 0.09333=0.1866$ gram atom of $H$.
Now,
$1$ gm atom of $H$ contains =$6.022\times {10}^{23}$ atoms of $H$.
$\therefore$ $0.1866$gram atom of hydrogen contains = $6.022\times {10}^{23}\times 0.1866=1.12\times {10}^{23}$ atoms
Conclusion: Number of $H$ atoms = $1.12\times {10}^{23}$ atoms