When 0 . 01mol of an organic compound containing 60 % carbon was...

Let $M is the molar mass of the compound \left(g / mol\right)$
mass of compound $= mole \times molecularmass = 0 . 01Mg$

mass of carbon $= 0 . 01M \times \dfrac{60}{100}$ ....(i)

moles of carbon $= \dfrac{massofcompound}{molecularmassofcarbon} = \dfrac{0 . 01M}{12} \times \dfrac{60}{100}$ ....(ii)

Thus, moles of ${CO}_{2}$ from combustion $= \dfrac{4 . 4}{44} =$0.1 moles of carbon dioxide

Using POAC,
Thus, from (i) and (ii),
$⇒ \dfrac{0 . 01M}{12} \times \dfrac{60}{100} = \dfrac{4 . 4}{44}$

$⇒ M = \dfrac{4 . 4}{44} \times \dfrac{100}{60} \times \dfrac{12}{0 . 01} = 200g / mol$