Q. If 5.85 g $NaCl\$(molecular weight 58.5) is dissolved in water and the solution is made up to 0.5 L, the molarity of the solution will be
A) 0.1
B) 0.2
C) 0.3
D) 0.4
Explanation
(b)
$M = \frac{W}{mol.\ wt. \times V(L)} = \frac{5.85}{58.5 \times 0.5} = 0.2\ M$
(a)6g of $NaOH/100\ mL$
(b)0.5 M $H_{2}SO_{4}$
$N = M \times basicity = 0.5 \times 2 = 1.0$
(c)$N$ phosphoric acid Normality=1
(d)$8\ g\ of\ KOH/L$
$Normality = \frac{strength\ in\ g/L}{equivalent\ weight} = \frac{8}{56} = 0.14\ N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.