Solutions and Colligative Properties - Result Question 54
####25. An organic compound $\left(C _x H _{2 y} O _y\right)$ was burnt with twice the amount of oxygen needed for complete combustion to $CO _2$ and $H _2 O$. The hot gases when cooled to $0^{\circ} C$ and $1 atm$ pressure, measured $2.24 L$. The water collected during cooling weight $0.9 g$. The vapour pressure of pure water at $20^{\circ} C$ is $17.5 mm Hg$ and is lowered by $0.104 mm$ when $50 g$ of the organic compound are dissolved in $1000 g$ of water. Give the molecular formula of the organic compound.
$(1983,5 M)$
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Solution:
- From lowering of vapour pressure information :
$$ \begin{array}{rlrl} & & \frac{0.104}{17.5}=\chi _2 & =\frac{n _2}{n _1+n _2} \ \Rightarrow & & \frac{n _1}{n _2}+1 & =168.27 \ \Rightarrow & \frac{n _1}{n _2} & =167.27 \ \Rightarrow & & \frac{1000}{18} \times \frac{M}{50} & =167.27 \ \Rightarrow & & M & =150 g / mol \end{array} $$
Also, the combustion reaction is :
$$ C _x H _{2 y} O _y+x O _2 \longrightarrow x CO _2+y H _2 O $$
$\because 18 y g$ of $H _2 O$ is produced from 1.0 mole of compound.
$\therefore 0.9 g$ of $H _2 O$ will be produced from $\frac{0.9}{18 y}=\frac{1}{20 y}$ mol
$\Rightarrow$ At the end, moles of $O _2$ left $=\frac{x}{20 y}$
moles of $CO _2$ formed $=\frac{x}{20 y}$
$\Rightarrow$ Total moles of gases at STP $=\frac{2 x}{20 y}=\frac{2.24}{22.4}$
$\Rightarrow \quad x=y$
$\Rightarrow$ Molar mass; $150=12 x+2 x+16 x=30 x$
$\Rightarrow \quad x=\frac{150}{30}=5$
$\Rightarrow \quad$ Formula $=C _5 H _{10} O _5$
