# A hydrocarbon contains $85.7 \%C$ If $42\;mg$ of the compound contains $3.01 \times 10^{20}$ molecules, the molecular formula of the compound is

$(a)\;C_6 H_{14} \\ (b)\;C_6H_{10} \\(c)\;C_6H_6 \\(d)\;C_6H_{12}$

$3.01 \times 10^{20}$ molecules are present in =$42 \times 10^{-3}\;g$ of hydrocarbon $(C_xH_y)$
$\therefore 6.023 \times 10^{23}$ molecules are present in
$\qquad= \large\frac{42 \times 10^{-3}}{3.01 \times 10^{20}}$$\times 6.023 \times 10^{23}$
$\qquad=84$
The molecular formula $=C_6H_{12}$
Hence d is the correct answer.