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 add is?

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

Answer 1

Answer: (C)

0.194 N

Answer 2

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 $NaOH =\frac{1}{40} \times 0.275\, eq .$ $=\frac{1}{40} \times 0.275 \times 1000$ $=6.88 \,meq$ $\therefore \underset{ (HCl)}{N_{1} V_{1}} = \underset{(NaOH)}{N_{2} V_{2}}$ $N_{1} \times 35.4 =6.88 (\because meq =N V)$ $N_{1} =0.194$