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

Related Formula

Core Logic

Let the initial volume of CO₂$CO_2$ be 1 dm³$1 \, \mathrm{dm^3}$. Let x$x$ be the volume of CO₂$CO_2$ reacted.

Reaction: CO₂(g) + C(s) arrow 2CO(g)$\quad CO_2(g) + C(s) \rightarrow 2CO(g)$ Initial: 1 dm³ 0$\quad 1 \, \mathrm{dm^3} \quad \quad \quad \quad 0$ Change: -x +2x$\quad -x \quad \quad \quad \quad +2x$ Eq / Final: 1 - x 2x$\quad 1 - x \quad \quad \quad \quad 2x$

Total volume of gaseous mixture after reaction = Volume of unreacted CO₂$CO_2$ + Volume of CO$CO$ formed

Step 1: Solve for x

Given Vtotal = 1.4 dm³$V_{\text{total}} = 1.4 \, \mathrm{dm^3}$ at STP. 1 + x = 1.4$1 + x = 1.4$

Step 2: Calculate Final Composition

Volume of CO₂$CO_2$ left = 1 - x = 1 - 0.4 = 0.6 dm³$= 1 - x = 1 - 0.4 = 0.6 \, \mathrm{dm^3}$ Volume of CO$CO$ formed = 2x = 2(0.4) = 0.8 dm³$= 2x = 2(0.4) = 0.8 \, \mathrm{dm^3}$

Pattern Recognition

When CO₂$CO_2$ reacts with solid Carbon to form 2CO$2CO$, every 1 mole (or volume) of CO₂$CO_2$ consumed results in a net increase of 1 mole (or volume) in the system. The total volume increase directly equals the volume of CO₂$CO_2$ reacted.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry