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$5ml$ of a gaseous hydrocarbon was exposed to $30ml$ of $O_2$.The resultant gas,on cooling is found to measure $25\;ml$ of which $10\;ml$ are obsorbed by NaOH and the remainder by pyrogallol.Determine molecular formula of hydrocarbons .All measurement are made at constant pressure and temperature.

$(a)\;CH_4\qquad(b)\;C_2H_4\qquad(c)\;C_2H_6\qquad(d)\;C_3H_8$

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Let formula of hydrocarbon be $C_aH_b$
$C_aH_b+[a+\large\frac{b}{4}]$$O_2\rightarrow aCO_2+\large\frac{b}{2}$$H_2O(l)$
Also given
Volume absorbed by $NaOHCO_2$ formed =10ml
Volume observed by pyragallol is of $O_2$ left=15ml.
Volume $O_2$ used =30-15=15ml
$5a=10$
$a=2$
$5(a+\large\frac{b}{4})$$=15$
$b=4$
$\therefore$ Hydrocarbon =$C_2H_4$
Hence (b) is the correct option.
answered Oct 30, 2013 by sreemathi.v
edited Mar 21, 2014 by balaji.thirumalai
 

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