Q. The amount of anhydrous $Na_{2}CO_{3}$ present in 250 $mL\$of 0.25 M solution is
A) 6.0 g
B) 6.625 g
C) 66.25 g
D) 6.225 g
Explanation
(b)
$M = \frac{w}{m \times V(L)}$
$0.25 = \frac{w}{106 \times 0.25}$
$\therefore w = 6.625\ g$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.