A hydrocarbon contains $10.5$ g of carbon per gram of hydrogen. $1$ litre of the hydrocarbon at $127^{0} C$ and $1$ atm pressure weighs $2.8$ g. Find the molecular formula.

C_{7} H_{8}

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C_{6} H_{10}

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C_{7} H_{6}

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C_{6} H_{12}

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Solution

The correct option is A $C_{7} H_{8}$
$1$ L at $127^{0} C$ corresponds to $\frac{273}{400}$ L at room temperature.
At STP, $1$ mole = $22.4$ L.
Thus, $\frac{273}{400}$ L $\times \frac{1 m o l}{22.4 L}$ = $0.0305$ mol = $2.8$ g
Molecular weight = $\frac{2.8 g}{0.0305 m o l} = 92 g m o l^{- 1}$ .
Hence, the molecular formula is $C_{7} H_{8}$ .

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A hydrocarbon contains 10.5 g of carbon per gram of hydrogen. 1 litre of the hydrocarbon at 1270C and 1 atm pressure weighs 2.8 g. Find the molecular formula.

The correct option is A C7H81 L at 1270C corresponds to 273400 L at room temperature.At STP, 1 mole = 22.4 L.Thus, 273400 L×1 mol22.4L = 0.0305 mol = 2.8 gMolecular weight = 2.8 g0.0305 mol=92gmol−1.Hence, the molecular formula is C7H8.

A hydrocarbon contains $10.5$ g of carbon per gram of hydrogen. $1$ litre of the hydrocarbon at $127^{0} C$ and $1$ atm pressure weighs $2.8$ g. Find the molecular formula.