When 1 dm3 of CO2 gas is passed over hot coke the volume of gaseous mixture after complete reaction at STP becomes 1.4…

Year	2024	Total
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2024

Total

Questions

Solution & Explanation

Related Formula

CO₂(g) + C(s) arrow 2CO(g)

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 = $(1 - x) + 2x = 1 + x = 1.4 dm³ x = 0.4 dm³$ .

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

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Q65 neet_2024_05_may_morning Mole Concept and Atom Calculations

The number of hydrogen atoms present in 5.4 g of urea is: (Given: Molar mass of urea:

60 g mol⁻¹

NA = 6.022 \times 10²³ particles mol⁻¹

)

A. (1) $2.168 \times 10²³$ $2.168 \times 10^{23}$
B. (2) $2.168 \times 10²²$ $2.168 \times 10^{22}$
C. (3) $1.084 \times 10²²$ $1.084 \times 10^{22}$
D. (4) $1.084 \times 10²³$ $1.084 \times 10^{23}$

Solution

Related Formula

Moles of urea = (w)/(M) = (5.4)/(60) = 0.09 mol

Formula of Urea = NH₂CONH₂ 4 H atoms per molecule

Core Logic

Number of H atoms = 0.09 \times 4 \times 6.022 \times 10²³ = 2.1679 \times 10²³ \approx 2.168 \times 10²³

Step 1: Conclusion

Number of hydrogen atoms is $2.168 \times 10²³$ $2.168 \times 10^{23}$.

Pattern Recognition

Urea contains 4 hydrogen atoms per molecule: $NH = 4 \times nurea \times NA$ $N_H = 4 \times n_{urea} \times N_A$.

Chapter Mix

Class 11 Chemistry: Some Basic Concepts of Chemistry

When $1 dm³$ $1 \text{ dm}^3$ of $CO₂$ $CO_2$ gas is passed over hot coke the volume of gaseous mixture after complete reaction at STP becomes $1.4 dm³$ $1.4 \text{ dm}^3$. The composition of the gaseous mixture at STP is:

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Selection

Pattern Recognition

Chapter Mix

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More Some Basic Concepts of Chemistry Previous-Year Questions

Solution

Related Formula

Core Logic

Step 1: Conclusion

Pattern Recognition

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When 1 dm³$1 \text{ dm}^3$ of CO₂$CO_2$ gas is passed over hot coke the volume of gaseous mixture after complete reaction at STP becomes 1.4 dm³$1.4 \text{ dm}^3$. The composition of the gaseous mixture at STP is:

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Selection

Pattern Recognition

Chapter Mix

Reference Study Guides

More Some Basic Concepts of Chemistry Previous-Year Questions

Solution

Related Formula

Core Logic

Step 1: Conclusion

Pattern Recognition

Chapter Mix

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When $1 dm³$ $1 \text{ dm}^3$ of $CO₂$ $CO_2$ gas is passed over hot coke the volume of gaseous mixture after complete reaction at STP becomes $1.4 dm³$ $1.4 \text{ dm}^3$. The composition of the gaseous mixture at STP is: