Q. Volume of 0.6 M NaOH required to neutralise $30\ {cm}^{3}$ of 0.4 M HCl is
A) $20\ {cm}^{3}$
B) $40\ {cm}^{3}$
C) $45\ {cm}^{3}$
D) $30\ {cm}^{3}$
Explanation
(a)
According to molarity equation
NaOH = HCl
${\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ M}_{1}V_{1\ } = \ M_{2}V_{2}$
$0.6\ \times V_{1\ }\ = \ 0.4\ \times \ 30\ \ \$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V_{1\ }\ = \ \frac{0.4\ \times \ 30\ }{0.6\ \ } = 20{cm}^{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.