Q. Calculate the number of atoms of hydrogen present in 5.6 g of urea, (NH2 )2 CO

Answer: H = \(2.248 \times 10^{23}\) atoms; N = \(1.124 \times 10^{23}\) atoms; C = \(5.62 \times 10^{22}\) atoms; O = \(5.62 \times 10^{22}\) atoms

Step 1: Molar mass of urea, (NH₂)₂CO

\(= 2(14) + 4(1) + 12 + 16 = 60\ \text{g mol}^{-1}\)

Step 2: Moles of urea

\(n = \dfrac{5.6}{60} = 0.0933\ \text{mol}\)

Step 3: Number of urea molecules

\(= 0.0933 \times 6.022 \times 10^{23} = 5.62 \times 10^{22}\) molecules

Step 4: Number of atoms (one molecule has 4 H, 2 N, 1 C, 1 O)

• H atoms \(= 4 \times 5.62 \times 10^{22} = 2.248 \times 10^{23}\)
• N atoms \(= 2 \times 5.62 \times 10^{22} = 1.124 \times 10^{23}\)
• C atoms \(= 1 \times 5.62 \times 10^{22} = 5.62 \times 10^{22}\)
• O atoms \(= 1 \times 5.62 \times 10^{22} = 5.62 \times 10^{22}\)