NEET 2026 (Phase 1) · ChemistryPrevious Year Question

When 1 dm $^{3}$ of $CO X_{2}$ gas is passed over hot coke, the volume of the gaseous mixture after complete reaction at STP becomes 1.4 dm $^{3}$ . The composition of the gaseous mixture at STP is

Answer: (B) 0.8 dm $^{3}$ $CO$ , 0.6 dm $^{3}$ $CO X_{2}$ . \textbf{Answer:} (B) 0.8 dm $^{3}$ $CO$ , 0.6 dm $^{3}$ $CO X_{2}$ \textbf{Solution:} $CO X_{2} + C (s) 2 CO$ .

A.0.8 dm $^{3}$ $CO$ , 0.8 dm $^{3}$ $CO X_{2}$
B.0.8 dm $^{3}$ $CO$ , 0.6 dm $^{3}$ $CO X_{2}$ ✓
C.0.6 dm $^{3}$ $CO$ , 0.8 dm $^{3}$ $CO X_{2}$
D.0.6 dm $^{3}$ $CO$ , 0.4 dm $^{3}$ $CO X_{2}$

Correct Answer

(B) 0.8 dm $^{3}$ $CO$ , 0.6 dm $^{3}$ $CO X_{2}$

Solution & Explanation

\textbf{Answer:} (B) 0.8 dm $^{3}$ $CO$ , 0.6 dm $^{3}$ $CO X_{2}$ \textbf{Solution:} $CO X_{2} + C (s) 2 CO$ . By Gay-Lussac's law, volumes behave like moles. Let $x$ dm $^{3}$ of $CO X_{2}$ react. Remaining $CO X_{2}$ $= 1 - x$ ; $CO$ formed $= 2 x$ . Total volume $= (1 - x) + 2 x = 1 + x = 1.4 \Rightarrow x = 0.4$ . $CO$ $= 2 (0.4) = 0.8$ dm $^{3}$ ; $CO X_{2}$ left $= 1 - 0.4 = 0.6$ dm $^{3}$ .

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When 1 dm3 of COX2​ gas is passed over hot coke, the volume of the gaseous mixture after complete reaction at STP becomes 1.4 dm3. The composition of the gaseous mixture at STP is

Solution & Explanation

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When 1 dm $^{3}$ of $CO X_{2}$ gas is passed over hot coke, the volume of the gaseous mixture after complete reaction at STP becomes 1.4 dm $^{3}$ . The composition of the gaseous mixture at STP is