Question

\(80\,\text{mL}\) of an organic compound is mixed with \(264\,\text{mL}\) of \(O_2\) and ignited. It gives \(224\,\text{mL}\) of gaseous mixture at NTP. After passing through KOH, \(64\,\text{mL}\) of gas remains. The organic compound is:

• A. \( \mathrm{C_2H_4} \)
• B. \( \mathrm{C_2H_2} \)
• C. \( \mathrm{C_4H_{10}} \)
• D. \( \mathrm{C_3H_6} \)

Hint:

For combustion volume problems:
KOH absorbs only \(CO_2\)
Volumes of gases are proportional to moles at same conditions
Use stoichiometry of combustion to find molecular formula

Answer 1

The Correct Option is B

Step 1: After combustion, total gaseous mixture \(= 224\,\text{mL}\). After passing through KOH (which absorbs \(CO_2\)): \[ \text{Remaining gas} = 64\,\text{mL} \Rightarrow \text{Unreacted } O_2 \] \[ \therefore CO_2 = 224 - 64 = 160\,\text{mL} \]
Step 2: Oxygen consumed: \[ O_2 \text{ used} = 264 - 64 = 200\,\text{mL} \]
Step 3: Let the compound be \( \mathrm{C_xH_y} \). From combustion: \[ 80\,\text{mL} \rightarrow 160\,\text{mL } CO_2 \Rightarrow x = 2 \]
Step 4: Oxygen required per mole: \[ \mathrm{C_xH_y} + \left(x + \frac{y}{4}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O \] \[ \frac{200}{80} = 2.5 = x + \frac{y}{4} \] \[ 2 + \frac{y}{4} = 2.5 \Rightarrow \frac{y}{4} = 0.5 \Rightarrow y = 2 \]
Step 5: Hence the compound is: \[ \boxed{\mathrm{C_2H_2}} \]