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 $



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