The number of hydrogen atoms present in $5.4\ \mathrm{g}$ of urea is:

(Given: Molar mass of urea : $60\ \mathrm{g\ mol^{-1}}$

$\mathrm{N_A}: 6.022 imes 10^{23}$ particles $\mathrm{mol^{-1}}$)

Question

The number of hydrogen atoms present in $5.4\ \mathrm{g}$ of urea is:

(Given: Molar mass of urea : $60\ \mathrm{g\ mol^{-1}}$

$\mathrm{N_A}: 6.022 	imes 10^{23}$ particles $\mathrm{mol^{-1}}$)

Accepted Answer

D. 1. **Recall the formula of urea and how many hydrogen atoms it contains.**

Urea is $\mathrm{NH_2CONH_2}$, also written as $\mathrm{CO(NH_2)_2}$. Counting the hydrogens in this formula, **each molecule of urea contains 4 hydrogen atoms**.

2. **Convert the given mass of urea into moles.**

To count atoms, we first need the number of moles of urea, using:

$$
	ext{moles} = \frac{	ext{given mass}}{	ext{molar mass}}
$$

Given mass is $5.4\ \mathrm{g}$ and molar mass is $60\ \mathrm{g\ mol^{-1}}$, so:

$$
	ext{moles of urea} = \frac{5.4}{60} = 0.09\ \mathrm{mol}
$$

3. **Convert moles of urea into moles of hydrogen atoms.**

Since each mole of urea contains $4$ moles of hydrogen atoms, we multiply:

$$
	ext{moles of H atoms} = 0.09 	imes 4 = 0.36\ \mathrm{mol}
$$

4. **Convert moles of hydrogen atoms into actual number of atoms using Avogadro's number.**

Avogadro's number tells us how many particles are present in one mole, so:

$$
	ext{Number of H atoms} = 0.36 	imes 6.022 	imes 10^{23} \approx 2.168 	imes 10^{23}
$$

Hence, the number of hydrogen atoms present is $2.168 	imes 10^{23}$, matching **Option D**.

The number of hydrogen atoms present in 5.4 g5.4\ \mathrm{g}5.4 g of urea is: (Given: Molar mass of urea : 60 g mol−160\ \mathrm{g\ mol^{-1}}60 g mol−1 NA:6.022×1023\mathrm{N_A}: 6.022 \times 10^{23}NA​:6.022×1023 particles mol−1\mathrm{mol^{-1}}mol−1)

The number of hydrogen atoms present in $5.4\ \mathrm{g}$ of urea is:

(Given: Molar mass of urea : $60\ \mathrm{g\ mol^{-1}}$

$\mathrm{N_A}: 6.022 \times 10^{23}$ particles $\mathrm{mol^{-1}}$ )