A 1.84 , mg sample of polyhydric alcoholic co | vedclass

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Question

(A) The reaction of a polyhydric alcohol with sodium or similar reagents releases $H_2$ gas based on the number of hydroxyl groups $(x)$: $R(OH)_x + x

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(A) The reaction of a polyhydric alcohol with sodium or similar reagents releases $H_2$ gas based on the number of hydroxyl groups $(x)$: $R(OH)_x + xNa ightarrow R(ONa)_x + \frac{x}{2} H_2$ Using the Principle of Atom Conservation $(PoAC)$ for hydrogen atoms in the $OH$ groups: $x 	imes 	ext{moles of } X = 2 	imes 	ext{moles of } H_2$ Given: Mass of $X = 1.84 \, mg = 1.84 	imes 10^{-3} \, g$ Molar mass of $X = 92.0 \, g/mol$ Volume of $H_2$ at $STP = 1.344 \, mL = 1.344 	imes 10^{-3} \, L$ Moles of $X = \frac{1.84 	imes 10^{-3}}{92} = 2 	imes 10^{-5} \, mol$ Moles of $H_2 = \frac{1.344 	imes 10^{-3}}{22.4} = 6 	imes 10^{-5} \, mol$ Substituting into the equation: $x 	imes (2 	imes 10^{-5}) = 2 	imes (6 	imes 10^{-5})$ $x = \frac{12 	imes 10^{-5}}{2 	imes 10^{-5}} = 6$ Thus,the number of alcoholic hydrogens is $6$.

List-$I$	List-$II$ (At $STP$)
$(A)$ $10 \ g \ CaCO_3 \xrightarrow{\Delta} \text{decomposition}$	$(i)$ $0.224 \ L \ CO_2$
$(B)$ $1.06 \ g \ Na_2CO_3 \xrightarrow{\text{Excess } HCl} \text{reaction}$	$(ii)$ $4.48 \ L \ CO_2$
$(C)$ $2.4 \ g \ C \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iii)$ $0.448 \ L \ CO_2$
$(D)$ $0.56 \ g \ CO \xrightarrow{\text{Excess } O_2} \text{combustion}$	$(iv)$ $2.24 \ L \ CO_2$
	$(v)$ $22.4 \ L \ CO_2$

$A$ $1.84 \, mg$ 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 $.....$

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