$0.5 \ L$ of gaseous hydrocarbon when burnt in excess of $O_2$ gave $3.5 \ L$ of $CO_2$ and $4 \ L$ of water vapours under same conditions. The molecular formula of the hydrocarbon will be:

  • A
    $C_4H_8$
  • B
    $C_4H_{10}$
  • C
    $C_7H_{16}$
  • D
    $C_5H_{12}$

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