Q. The volume of water to be added to $\frac{N}{2}$ HCl to prepare $500\ {cm}^{3}$ of $\frac{N}{10}$ solution is
A) $450\ {cm}^{3}$
B) $100\ {cm}^{3}$
C) $45\ {cm}^{3}$
D) $400\ {cm}^{3}$
Explanation
(d)
Let x mL of HCl are taken , then
$N_{1}V_{1}\ =$ $N_{2}V_{2}$
$\frac{1}{2} \times x = \frac{1}{10} \times 500$
x = 100mL
Hence, water needed to add
= 500-100= 400mL
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.