The number of hydrogen atoms present in 5.4 g of urea is: (Given: Molar mass of urea: 60 g mol⁻¹$60 \text{ g mol}^{-1}$, NA = 6.022 × 10²³ particles mol⁻¹$N_A = 6.022 \times 10^{23} \text{ particles mol}^{-1}$)

Related Formula

Core Logic

Initial volume of CO₂ = 1 dm³$CO_2 = 1 \text{ dm}^3$. At equilibrium: Volume of CO₂ = 1 - x$CO_2 = 1 - x$ Volume of CO = 2x$CO = 2x$ Total volume = (1 - x) + 2x = 1 + x = 1.4 dm³ x = 0.4 dm³$(1 - x) + 2x = 1 + x = 1.4 \text{ dm}^3 \implies x = 0.4 \text{ dm}^3$.

Volume of CO₂ = 1 - 0.4 = 0.6 dm³$CO_2 = 1 - 0.4 = 0.6 \text{ dm}^3$. Volume of CO = 2(0.4) = 0.8 dm³$CO = 2(0.4) = 0.8 \text{ dm}^3$.

Step 1: Final Selection

Composition is 0.8 dm³$0.8 \text{ dm}^3$ of CO and 0.6 dm³$0.6 \text{ dm}^3$ of CO₂$CO_2$.

Pattern Recognition

1 mole of CO₂$CO_2$ produces 2 moles of CO. Expansion in volume Δ V = x$\Delta V = x$.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry