The boiling point elevation of 0.30g acetic acid in 100g benzene is 0.0633K.Calculate the molar mass of acetic acid from this data?What conclusion can you draw about the molecular state of the solute in the solution?[Given Kb for benzene = 2.53Kkgmol−1]
Answers
Answered by
33
Here
W1=100g
W2=0.30g
ΔTb=0.0633K
Kb=2.53Kkgmol−1
∴ Molar mass of acetic acid M2=Kb×W2×1000/ΔTb×W1
⇒2.53×0.30×1000/0.0633×100
⇒119.90
Van't Hoff factor
i=Normal molar mass/Abnormal molar mass
=60/119.90
=0.5
Hence i<1 ,therefore the solute (acetic acid) is associated.
W1=100g
W2=0.30g
ΔTb=0.0633K
Kb=2.53Kkgmol−1
∴ Molar mass of acetic acid M2=Kb×W2×1000/ΔTb×W1
⇒2.53×0.30×1000/0.0633×100
⇒119.90
Van't Hoff factor
i=Normal molar mass/Abnormal molar mass
=60/119.90
=0.5
Hence i<1 ,therefore the solute (acetic acid) is associated.
Answered by
0
Answer:
we can easily find molecular weight of acetic acid(CH3COOH) by the given data
Attachments:
Similar questions