Q. 3.0 molal $NaOH$ solution has a density of 1.110 g/$mL$. The molarity of the solution is
A) 3.9732
B) 2.9732
C) 1.9732
D) 0.9732
Explanation
(b)
$Molarity(m) = \frac{M}{1000d - MM^{'}} \times 1000$
Where $M^{'} =$molar mass of solute
$3 = \frac{M \times 1000}{1000 \times 1.11 - M \times 40}$
$1000M = 3330 - 120\ M$
$1120\ M = 3330$
$M = \frac{3330}{1120} = 2.9732$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.