A 1.84 mg sample of polyhydric alcoholic compound ' X ' of molar mass 92.0 g / mol gave 1.344 mL of H 2 gas at STP. The number of alcoholic hydrogens present in compound ' X ' is

Question

A 1.84 mg sample of polyhydric alcoholic compound ' X ' of molar mass 92.0 g / mol gave 1.344 mL of H 2 gas at STP. The number of alcoholic hydrogens present in compound ' X ' is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

R ( OH ) x arrow H 2 PoAC on H - x((1.84 × 10-3/92))=(1.344/22.4) × 2 x=(1.344 × 2 × 92 × 1000/1.84 × 22400)=6 x=6

Q. A $1.84 m g$ sample of polyhydric alcoholic compound ' $X$ ' of molar mass $92.0 g / m o l$ gave $1.344 m L$ of $H_{2}$ gas at STP. The number of alcoholic hydrogens present in compound ' $X$ ' is

$R (O H)_{x} \to H_{2}$
PoAC on $H -$
$x (\frac{1.84 \times 1 0 ^{- 3}}{92}) = \frac{1.344}{22.4} \times 2$
$x = \frac{1.344 \times 2 \times 92 \times 1000}{1.84 \times 22400} = 6$
$x = 6$

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

R(OH)x​→H2​ PoAC on H− x(921.84×10−3​)=22.41.344​×2 x=1.84×224001.344×2×92×1000​=6 x=6

Q. A $1.84 m g$ sample of polyhydric alcoholic compound ' $X$ ' of molar mass $92.0 g / m o l$ gave $1.344 m L$ of $H_{2}$ gas at STP. The number of alcoholic hydrogens present in compound ' $X$ ' is

$R (O H)_{x} \to H_{2}$
PoAC on $H -$
$x (\frac{1.84 \times 1 0 ^{- 3}}{92}) = \frac{1.344}{22.4} \times 2$
$x = \frac{1.344 \times 2 \times 92 \times 1000}{1.84 \times 22400} = 6$
$x = 6$