Q. Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is
A) 1.14 mol ${kg}^{- 1}$
B) 3.28 mol ${kg}^{- 1}$
C) 2.28 mol ${kg}^{- 1}$
D) 0.44 mol ${kg}^{- 1}$
Explanation
(c)
Molality (m) =$\frac{M}{1000d - MM_{1}} \times 100$
M = Molarity
$M_{1}$ = Molecular mass
d = density
=$\frac{2.05}{(1000\ \times \ 1.02) - (2.05\ \times \ 60)} \times$100
=2.28 mol ${kg}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.