Q. 35.4 mL of HCl is required for the neutralisation 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
Explanation
(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 = NV)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ N_{1} = 0.194$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.