The volume of HCl,containing 73 g L^{-1},re | vedclass

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Question

(C) $1$. Calculate moles of $Na$: $n(Na) = \frac{0.69 \ g}{23 \ g \ mol^{-1}} = 0.03 \ mol$.$2$. Reaction with water: $Na + H_2O \longrightarrow NaOH

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$15$

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(C) $1$. Calculate moles of $Na$: $n(Na) = \frac{0.69 \ g}{23 \ g \ mol^{-1}} = 0.03 \ mol$.$2$. Reaction with water: $Na + H_2O \longrightarrow NaOH + \frac{1}{2} H_2$.$3$. From the stoichiometry,$1 \ mol$ of $Na$ produces $1 \ mol$ of $NaOH$. Thus,$n(NaOH) = 0.03 \ mol$.$4$. Neutralization reaction: $NaOH + HCl \longrightarrow NaCl + H_2O$.$5$. Moles of $HCl$ required = $0.03 \ mol$.$6$. Concentration of $HCl$ solution: Molarity $M = \frac{73 \ g \ L^{-1}}{36.5 \ g \ mol^{-1}} = 2 \ M$.$7$. Volume of $HCl$ required: $V = \frac{n}{M} = \frac{0.03 \ mol}{2 \ mol \ L^{-1}} = 0.015 \ L = 15 \ mL$.

The volume of $HCl$,containing $73 \ g \ L^{-1}$,required to completely neutralize $NaOH$ obtained by reacting $0.69 \ g$ of metallic sodium with water,is $........ mL$. (Nearest Integer) (Given: molar masses of $Na, Cl, O, H$ are $23, 35.5, 16$ and $1 \ g \ mol^{-1}$ respectively)

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