A 1 . 84mg sample of polyhydric alcoholic compound ' X ' of molar...

Number of moles of alcoholic compound $= \dfrac{Givenmass}{Molarmass} = \dfrac{1 . 84 \times {10}^{- 3}}{92 . 0} = 2 \times {10}^{- 5}moles$

Number of moles of hydrogen gas produced $= \dfrac{VolumeofgasatSTP}{22400ml} = \dfrac{1 . 344}{22400} = 6 \times {10}^{- 5}moles$

Monohydric alcohol will react with active metal to give $\dfrac{1}{2}mole$ of hydrogen gas.

Since per mole of given alcohol is giving 3 moles of hydrogen gas.

So number of alcoholic hydrogen present in compound should be =6.