Q. 9.8 g of $H_{2}SO_{4}$ is present in 2 L of a solution. The molarity of the solution is
A) 0.05 M
B) 0.01 M
C) 0.03 M
D) 0.02 M
Explanation
(a)
$M = \frac{w \times 1000}{mol.\ mass \times volume\ in\ mL}$
$= \frac{9.8 \times 1000}{98 \times 2000} = 0.05\ M$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.