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प्रश्न
3.9 g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point of 1.62 K. Calculate the van't Hoff factor and predict the nature of solute (associated or dissociated).
(Given : Molar mass of benzoic acid = 122 g mol−1, Kf for benzene = 4.9 K kg mol−1)
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उत्तर
We know that the depression in freezing point is given by
`DeltaT_`
Here,
van't Hoff factor = i
Depression in freezing point, ΔTf=1.62 K
Kf for benzene=4.9 K kg mol−1
Mass of benzoic acid, ws=3.9 g
Mass of benzene, W=49 g
Molar mass of benzoic acid, Ms=122 g mol−1
Substituting the values, we get
`1.62 = (ixx4.9xx(3.9xx1000))/(122xx49)`
`rArri=(1.62xx122xx49)/(4.9xx3.9xx1000)`
= 0.51
As the value of i < 1, benzoic acid is an associated solute.
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