An organic compound (Cx H2y Oy) was burnt with twice the amount of oxygen needed for complete combustion to CO2 and H2O. The hot gases when cooled to 0°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°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.

Question

An organic compound (Cx H2y Oy) was burnt with twice the amount of oxygen needed for complete combustion to CO2 and H2O. The hot gases when cooled to 0°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°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. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

From lowering of vapour pressure information :

\frac{0.104}{17.5} = χ_{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{100}{18} \times \frac{M}{50} = 167.27

\Rightarrow M = 150 g / m o l

Also, the combustion reaction is :

C_{x} H_{2 y} O_{y} + x O_{2} \to x C O_{2} + y H_{2} O

∵ 18 y g

of

H_{2} O

is produced from

1.0

mole of compound.

∴ 0.9 g

of

H_{2} O

will be produced from

\frac{0.9}{18 y} = \frac{1}{20 y} m o l

\Rightarrow

At the end, moles of

O_{2}

left =

\frac{x}{20 y}

moles of

C O_{2}

formed

= \frac{x}{20 y}

\Rightarrow

Total moles of gases at STP =

\frac{2 x}{20 y} = \frac{2.24}{22.4}

\Rightarrow x = y

\Rightarrow

Molar mass;

150 = 12 x + 2 x + 16 x = 30 x

\Rightarrow x = \frac{150}{30} = 5

\Rightarrow

Formula

= C_{5} H_{10} O_{5}