Solutions and Colligative Properties - Result Question 25
25. An organic compound $\left(\mathrm{C}x \mathrm{H}{2 y} \mathrm{O}_y\right)$ was burnt with twice the amount of oxygen needed for complete combustion to $\mathrm{CO}_2$ and $\mathrm{H}_2 \mathrm{O}$. The hot gases when cooled to $0^{\circ} \mathrm{C}$ and $1$ atm pressure, measured $2.24$ L . The water collected during cooling weight $0.9$ g . The vapour pressure of pure water at $20^{\circ} \mathrm{C}$ 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 :
$\dfrac{0.104}{17.5}=\chi_2=\dfrac{n_2}{n_1+n_2}$
$\Rightarrow \quad \dfrac{n_1}{n_2}+1=168.27$
$\Rightarrow \quad \dfrac{n_1}{n_2}=167.27$
$\Rightarrow \quad \dfrac{1000}{18} \times \dfrac{M}{50}=167.27$
$\Rightarrow \quad M=150 \mathrm{~g} / \mathrm{mol}$
Also, the combustion reaction is :
$ \mathrm{C}x \mathrm{H}{2 y} \mathrm{O}_y+x \mathrm{O}_2 \longrightarrow x \mathrm{CO}_2+y \mathrm{H}_2 \mathrm{O} $
$\because 18 y \mathrm{~g}$ of $\mathrm{H}_2 \mathrm{O}$ is produced from $1.0$ mole of compound.
$\therefore 0.9 \mathrm{~g}$ of $\mathrm{H}_2 \mathrm{O}$ will be produced from $\dfrac{0.9}{18 y}=\dfrac{1}{20 y} \mathrm{~mol}$
$\Rightarrow \quad$ At the end, moles of $\mathrm{O}_2$ left $=\dfrac{x}{20 y}$
moles of $\mathrm{CO}_2$ formed $=\dfrac{x}{20 y}$
$\Rightarrow \quad$ Total moles of gases at STP $=\dfrac{2 x}{20 y}=\dfrac{2.24}{22.4}$
$\Rightarrow \quad x=y$
$\Rightarrow \quad$ Molar mass; $150=12 x+2 x+16 x=30 x$
$\Rightarrow \quad x=\dfrac{150}{30}=5$
$\Rightarrow \quad$ Formula $=\mathrm{C}5 \mathrm{H}{10} \mathrm{O}_5$
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