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 = 1.84*10−392=2.0*10−5mole. Mole of H2 gas produced = 1.34422400=6.0*10−5mole. No of OH gp present = 6.0*10−52.0*10−5=3

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_________.

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Payal Gupta | Contributor-Level 10

7 months ago

Mole of polyhydric alcohol = $\frac{1.84 * 1 0^{- 3}}{92} = 2.0 * 1 0^{- 5} m o l e .$

Mole of H₂ gas produced = $\frac{1.344}{22400} = 6.0 * 1 0^{- 5} m o l e .$

No of OH gp present = $\frac{6.0 * 1 0^{- 5}}{2.0 * 1 0^{- 5}} = 3$

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