Question

12 ml of a gaseous mixture of an alkane and an alkene (containing same number of carbon atoms) require exactly 285 ml of air (containing $20\%$ by volume of $O_2$ and rest $N_2$) for complete combustion at room temperature. After combustion when the gaseous mixture is passed through a KOH solution, it shows a volume contraction of 36 ml. Find the true statement(s) in the following options:

• A. The alkane is propane
• B. The alkene is butene
• C. Mole fraction of $C{O}_2$ in the final gaseous mixture is $\frac{3}{22}$
• D. Liquid water formed is 42 ml

Answer 1

Answer: (A), (C)

Mole fraction of $C{O}_2$ in the final gaseous mixture is $\frac{3}{22}$

Let the alkane be $C_n{H}_{2n+2}$ and alkene be $C_n{H}_{2n}$ and let $x$ ml be the volume of the alkane and $12-x$ ml be the volume of alkene.

$C_n{H}_{2n+2}\left(g\right)+\frac{3n+1}{2}{O}_2\left(g\right)\to nC{O}_2\left(g\right)+(n+1)H_{2}O\left(l\right)$
$C_n{H}_{2n}\left(g\right)+\frac{3n}{2}{O}_2\left(g\right)\to nC{O}_2\left(g\right)+n{H}_{2}O\left(l\right)$

Volume of $C{O}_2$ produced by combustion of alkane = $nx$ ml
Volume of $C{O}_2$ produced by combustion of alkene = $n(12-x)$ ml

The contraction of gaseous mixture in KOH is due to $C{O}_2$, i.e. vol of $C{O}_2$ is 36 ml.
So, $nx+n(12-x)=36\Rightarrow n=3$
a) The alkane is propane.
b) The alkene is propene.
According to the question,
$N_2$ in 285 ml air $=285\times 0.80=228ml$
c) So, mole fraction of $C{O}_2$ in final gaseous mixture $=\frac{36}{36+228}=\frac{36}{264}=\frac{3}{22}$
d) The volume of oxygen is $285\times \frac{20}{100}=57ml$
$\therefore x\times \frac{3n+1}{2}+(12-x)\times \frac{3n}{2}=57$
Putting, $n=3$ we get, $x=6$.
So, 6 ml of propane and 6 ml of propene are present in the mixture. So moles of liquid water formed,
$\frac{6mL\times (n+1+n)mol}{22.4L}=\frac{6\times 7}{22.4}mmol$
Therefore mass of liquid water will be,
$\frac{6\times 7}{22.4}mmol\times 18g/mol=33.75mg$
So, volume of liquid water will be $0.03375ml$ because the density of water at room temperature is 1 g/mL.