A 1.84 mg sample of polyhydric alcoholic compound 'X' of molar mass 92.0 g/mol gave1.344 mL of H₂
gas at STP. The number of alcoholic hydrogens present in compound 'X'is____. 2

volume of hydrogen gas = 1.344ml

mole of hydrogen gas = $\frac{1.344}{22400}=6 x {10}^{-5}$

number of hydrogen atoms per molecule of H₂ = 2

moles of H atoms produced = 12 x 10 ^-5

No. of moles of organic compound = $\frac{1.84x{10}^{-3}}{92}=2 x {10}^{-5}$ 

No. of $-OH$ (hydroxyl group in one molecule)= $\frac{12x{10}^{-5}}{2x{10}^{-5}}=6$