100 ml of a hydrocarbon on combustion gave $300 m l o f C O_{2}$ and 400 ml of water vapour, measured under similar conditions of temperature and pressure. The molecular formula of hydrocarbon is:

C_{3} H_{6}

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C_{3} H_{4}

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C_{3} H_{8}

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C H_{4}

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Solution

The correct option is C $C_{3} H_{8}$
For combustion reaction,
$C_{x} H_{y} + (x + \frac{y}{4}) O_{2} \to x C O_{2} + \frac{y}{2} H_{2} O$
$100 m l$ $300 m l$ $400 m l$

Under same pressure and temperature conditions, volume is proportional to moles.
$\Rightarrow$ 1 mole of hydrocarbon gives 3 moles of $C O_{2}$ and 4 moles of $H_{2} O$ .
$\Rightarrow x = 3; \frac{y}{2} = 4 o r y = 8$

Hence, the formula is
$C_{x} H_{y} ⟶ C_{3} H_{8}$

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100 ml of a hydrocarbon on combustion gave 300mlofCO2 and 400 ml of water vapour, measured under similar conditions of temperature and pressure. The molecular formula of hydrocarbon is:

The correct option is C C3H8For combustion reaction,CxHy+(x+y4)O2→xCO2+y2H2O100ml 300ml 400mlUnder same pressure and temperature conditions, volume is proportional to moles. ⇒ 1 mole of hydrocarbon gives 3 moles of CO2 and 4 moles of H2O.⇒x=3; y2=4ory=8Hence, the formula is CxHy⟶C3H8

100 ml of a hydrocarbon on combustion gave $300 m l o f C O_{2}$ and 400 ml of water vapour, measured under similar conditions of temperature and pressure. The molecular formula of hydrocarbon is: