Question

$14.96 \,g$ of a mixture containing $n$-hexane and ethyl alcohol are reacted with sodium metal to give $200\, ml$ of $H _{2}$ at $27^{\circ} C$ and $760 \,mm$ pressure. The percentage of ethyl alcohol in the mixture is ______. $[ R =0.082\, L$ $\left. atm\, mol ^{-1} K ^{-1}\right]$

Answer 1

Only ethyl alcohol reacts with sodium metal.
$C _{2} H _{5} OH + Na \rightarrow C _{2} H _{5} ONa +1 / 2 H _{2}$
No. of moles of $H _{2}$ formed $=\frac{ PV }{ RT }$
$=\frac{760}{760} \times \frac{200}{1000} \times \frac{1}{0.082 \times 300} $
$=\frac{1}{15 \times 8.2}-\frac{1}{123}-8.13 \times 10^{-3}$
$0.5$ mole of $H _{2}$ is liberated by $1$ mole of $C _{2} H _{5} OH$.
$8.13 \times 10^{-3}$ mole $H _{2}$ is liberated by
$\frac{1 \times 8.13 \times 10^{-3}}{0.5}$ mole of $C _{2} H _{5} OH$
$=16.26 \times 10^{-3}$ mole of $ C _{2} H _{3} OH $
$=16.26 \times 10^{-3} \times 46 \,g$ of $C _{2} H _{5} OH $
$=0.748 \,g$
The percentage of ethyl alcohol $n$ the mixture
$=\frac{7.48 \times 100}{14.96}=\frac{74.8}{14.96}=5 \%$