An organic compound $C_{x} H_{2 y} O_{y}$ was burnt with twice the amount of $O_{2}$ gas as required for complete combustion into ${CO}_{2} and H_{2} O$ . The hot gases when cooled to $0^{0} C$ and 1 atm pressure measured 2.24 litres. The water collected during the cooling weighs 0.9 grams. The vapour pressure of pure water is 17.5 mm of Hg at $20^{0} C$ and is lowered by 0.104 mm of Hg when 50 grams of organic compound is dissolved in 1000 grams water.(Assume that organic compound is non volatile and does not dissociate or associate). The molecular formula of the compound is

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$C_{5} H_{10} O_{5}$

$C_{10} H_{10} O_{5}$

$C_{5} H_{5} O_{10}$

$C_{5} H_{10} O_{10}$

answer is A.

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Detailed Solution

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

Number of moles after cooling =2x
Volume after cooling=2.24 litres
Number moles of $C O_{2}$ =0.05

$∴$ EF= $C H_{2} O$

$P^{0} = 17.5 m m$

$P^{0} - P = 0.104 m m$

M=151.4
MF= $C_{5} H_{10} O_{5}$

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