Q. 19.85 mL of 0.1 N $NaOH$ reacts with 20 mL of HCl solution for complete neutralization. The molarity of HCl solution is
A) 9.9
B) 0.99
C) 0.099
D) 0.0099
Explanation
(c)
$Molarity\ of\ base = \frac{Normality}{Acidity} = \frac{0.1}{1} = 0.1$
$M_{1}V_{1} = M_{2}V_{2}$
$0.1 \times 19.85 = M_{2} \times 20$
$M_{2} = 0.09925 \approx 0.099$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.