Question

# The freezing point of benzene decreases by 0.45∘C when 0.2g of acetic acid is added to 20g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be (Kf for benzene = 5.12K kg mol−1)

A
64.6%
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
80.4%
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
74.6%
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
94.6%
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## The correct option is D 94.6%Let the degree of association of acetic acid (CH3COOH) in benzene is α, then 2CH3COOH⇌(CH3COOH)2 Initial moles10Moles at equilibrium1−αα2 ∴ Total moles=1−α+α2=1−α2 or i=1−α2 Now, depression in freezing point (ΔTf) is given as ΔTf=i Kfm . . . . (i) Where, Kf - molal depression constant or cryoscopic constant. m = Molality Molality=number of moles of soluteweight of solvent (in kg)=0.260×10020 Putting the values in Eq.(i) ∴ 0.45=[1−α2] (5.12)[100020] 1−α2=0.45×60×205.12×0.2×1000 ⇒1−α2=0.527 ⇒α2=1−0.527 ∴ α=0.946 Thus, percentage of association = 94.6%

Suggest Corrections
14
Join BYJU'S Learning Program
Select...
Related Videos
Abnormal Colligative Properties
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
Select...