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JEE Main 20163 April
Exam
JEE Main
Year
2016
Date
3 April
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry

JEE Main 2016 · 3 April · Chemistry · Some Basic Concepts of Chemistry

At 300K and 1atm , 15mL of a gaseous hydrocarbon requires 375mL air...

At 300K300K and 1atm1atm, 15mL15mL of a gaseous hydrocarbon requires 375mL375mL air containing 2020 % {O}_{2} by volume, for complete combustion. After combustion, the gases occupy 345mL345mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is:
(Assume complete combustion of reactant)

Options

  1. AC4H8{C}_{4}{H}_{8}
  2. BC4H10{C}_{4}{H}_{10}
  3. CC3H6{C}_{3}{H}_{6}
  4. DC3H8{C}_{3}{H}_{8}

Correct answer

D. C3H8{C}_{3}{H}_{8}

Step-by-step solution

Volume of N2{N}_{2} in air=375×0.8=300ml= 375 \times 0.8 = 300ml
Volume of O2{O}_{2} in air=375×0.2=75ml= 375 \times 0.2 = 75ml
{CxHy+(x+y4)O2⟶xCO2(g)+y2H2O(ℓ)15ml15(x+y4)0015x−\begin{cases}{\text{C}}_{\text{x}} {\text{H}}_{\text{y}} & + & \left(\text{x} + \dfrac{\text{y}}{4}\right) {\text{O}}_{2} & ⟶ & {\text{xCO}}_{2} \left(\text{g}\right) & + & \dfrac{\text{y}}{2} {\text{H}}_{2} \text{O} \left(ℓ\right) \\ 1 5 \text{ml} & & 1 5 \left(\text{x} + \dfrac{\text{y}}{4}\right) & & & & \\ 0 & & 0 & & 1 5 \text{x} & & -\end{cases}
After combustion, total volume
345=VN2+VCO2345 = {V}_{{N}_{2}} + {V}_{{CO}_{2}}
345=300+15x345 = 300 + 15x
x=3x = 3
Volume of O2{O}_{2} used
15(x+y4)=7515\left(x + \dfrac{y}{4}\right) = 75
x+y4=5x + \dfrac{y}{4} = 5
y=8y = 8
So, hydrocarbon isC3H8{C}_{3}{H}_{8}.
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