How much of NaOH (in g) is required to neutra | vedclass

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(B) The neutralization reaction is $NaOH + HCl \rightarrow NaCl + H_2O$.For neutralization,the number of equivalents of $NaOH$ must equal the number o

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$6$

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(B) The neutralization reaction is $NaOH + HCl ightarrow NaCl + H_2O$.For neutralization,the number of equivalents of $NaOH$ must equal the number of equivalents of $HCl$.Number of equivalents of $HCl = N 	imes V 	ext{ (in } L) = 0.1 \ N 	imes 1.5 \ L = 0.15 \ 	ext{equivalents}$.Since the $n$-factor of $NaOH$ is $1$,its equivalent weight is equal to its molar mass,which is $23 + 16 + 1 = 40 \ g/mol$.Mass of $NaOH = 	ext{Number of equivalents} 	imes 	ext{Equivalent weight} = 0.15 	imes 40 = 6 \ g$.

How much of $NaOH$ (in $g$) is required to neutralize $1500 \ cm^3$ of $0.1 \ N \ HCl$? (Given: $Na = 23, O = 16, H = 1$)

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