JEE Mains · Chemistry · STD 11 - 1. Some basic concept of chemistry
80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to 273 K occupy 224 mL. When the system is treated with KOH solution, the volume decreases to 64 mL. The formula of the hydrocarbon is:
- A \( C_{2}H_{4} \)
- B \( C_{4}H_{10} \)
- C \( C_{2}H_{2} \)
- D \( C_{2}H_{6} \)
Answer & Solution
Correct Answer
(C) \( C_{2}H_{2} \)
Step-by-step Solution
Detailed explanation
\(C _{ x } H _{ y ( g )}+\left( x +\frac{ y }{4}\right) O _{2(g)} \longrightarrow xCO _{2(g)}+\frac{ y }{2} H _2 O _{(\ell)}\) \(\begin{array}{lllll} t =0 & 80 & 264 & 0\end{array}-\) \(t = t _{\text {final }} \quad-\quad 264-80\left( x +\frac{ y }{4}\right) \quad 80 x\)…
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