Some Basic Concepts Of Chemistry Question 9
Question: Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is
Options:
A) 0.44 mol $ k{g^{-1}} $
B) 1.14 mol $ k{g^{-1}} $
C) 3.28 mol $ k{g^{-1}} $
D) 2.28 mol $ k{g^{-1}} $
Show Answer
Answer:
Correct Answer: D
Solution:
weight of acetic acid = $ 2.05\times 60=123 $ weight of solution = $ 1000\times 1.02=1020 $
$ \therefore $ weight of water = (1020-123)=897 g Molality = $ \frac{moles,of,solute}{kg,of,solvent} $
$ \therefore $ Molality = $ \frac{2.05\times 1000}{897}=2.285 $
