Density of a 2.05M solution of acetic acid in water is 1.02g/ml.find molality of solution is===== tnk u
Answer (1)
5th Sep, 2020
Hello,
Let the volume of the solution is 1 L or 1000 ml
Now, 1 ml = 1.2 gm, therefore 1000 ml = 1000 * 1.2 = 1200 gm
Number of moles in 1 L of solution = 2.05 moles
Mass of solute = nunmber of moles * molecular weight
= 2.05 * 60 = 123 gm
Solvent mass = 1020 - 123 = 897 gm
Now we know that Molality = moles of solute/kg of solvent
Molality = 123/60 * 1000/897
Solving the above equation, we get:
Molality = 2.28 mol/kg (Answer)
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