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 \[k{{g}^{-1}}\]
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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