A 1.84 mg sample of polyhydric alcoholic compound 'X' of molar mass 92.0 g/mol gave 1.344mL of H₂ gas at STP. The number of alcoholic hydrogens present in compound 'X' is_________.

Mole of polyhydric alcohol =  $\frac{1.84*10^{-3}}{92}=2.0*10^{-5}mole.$

Mole of H₂ gas produced = $\frac{1.344}{22400}=6.0*10^{-5}mole.$  

No of -OH gp present =  $\frac{6.0*10^{-5}}{2.0*10^{-5}}=3$