Question

$35.4 \,mL$ of $ HCl $ is required for the neutralization of a solution containing $0.275\, g$ of sodium hydroxide. The normality of hydrochloric acid is?

• A. 0.97 N
• B. 0.142 N
• C. 0.194 N
• D. 0.244 N

Answer 1

Answer: (C)

We know that $1 \,g$ equivalent weight of
$ NaOH =40_{g} $
$ \therefore 40 \,g$ of $ NaOH =1 \,g $ eq. of $NaOH $
$\therefore 0.275 \,g$ of $NaOH =\frac{1}{40} \times 0.275 \,eq$
$=\frac{1}{40} \times 0.275 \times 1000$
$=6.88$ meq
$\therefore N_{1} V_{1}=N_{2} V_{2}$
$( HCl ) \,\,\,\,( NaOH )$
$N_{1} \times 35.4=6.88 $
$(\because meq =N V)$
$N_{1}=0.194$