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3.9 g of benzoic acid dissolved in 49 g ...

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), K_f` for benzene `="4.9 K kg mol"^(-1)`

Text Solution

Verified by Experts

The given quantities are :
`W_(2) =3.8g, W_(1)=49g`
`triangleT_(F)=1.62 K`,
`K_(f)=4.9"K kg mol"^(-1)`
Apply van.t Hoff equation
`triangleT_(f)=Ik_(f).m`
where, i=number of ions
`triangleT_(f)=(i.K_(f) xx W_(2) xx 10000)/(M_(2) xx W_(1))`
`rArr 1.62 K =(i.4.9"K kg mol"^(-1) xx 3.9g xx 1000)/("122 g mol"^(-1) xx 49g)`
`i=(1.62 xx 122 xx 49)/(4.9 xx 3.9 1000)=0.51`
As the value of `i lt 1` thus benoic acid is an associated solute.
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