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5 mL of a gaseous hydrocarbon was expose...

`5 mL` of a gaseous hydrocarbon was exposed to `30 mL` of `O_(2)`. The resultant gas, on cooling, is formed to measure `25 mL` of which `10 mL` is absrobed by `NaOH` and the remainder by pyrogallol. Determine the molecular formula of hydrocation. All measurements are made at constant pressure and temperature.

Text Solution

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Let the formula of hydrocarbon `= C_(x) H_(y)`
`:. C_(x) H_(y) + (x + (y)/(4)) O_(2) rarr xCO_(2) + (y)/(2) H_(2) O`
`"Initial" 1 mL (x + (y)/(4)) mL 0 0` ltbgt `"Final" 5 mL 5 (x + (y)/(4)) mL 5x mLm -`
`CO_(2) = 10 mL`
`:. 5 x = 10 implies x = 2 mL`
Volume of `O_(2)` left `= 15 mL`
Volume of `O_(2)` used `= 30 - 15 = 15 mL`.
`:. 5 (x + (y)/(4)) = 15 implies y = 4`
Formula of hydrocarbon `= C_(2) H_(4)`
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