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The freezing point of a solution contain...

The freezing point of a solution containing 5g of benzoic acid `(M=122 g mol^(-1))` in 35g of benzene is depressed by 2.94 K. What is the percentage association of benzoic acid if it forms a dimer in solution. [`K_(f)` for benzene = 4.9 K kg `mol^(-1)` ]

Text Solution

Verified by Experts

Number of moles of benzoic acid `=(5)/(122)`
Molality of benzoic acid solution `= (5)/(122) xx (1000)/(35) =1.17`
Apply the relation
`Delta T_(f) =i K_(f) m`, where i is van.t Hofff factor
`2.94 K=ixx4.8 Kg mol^(-1) xx 1.17 mol kg^(-1)`,
or `i=(2.94)/(4.9 xx 1.17)" "....(i)`
`(C_(6) H_(5) COOH)_(2) to (C_(6) H_(5) COOH)_(2)`
Total number of moles `=1-x (x//2) =1 - (x//2)`
`i= (1-(x)/(2))/(1)" "...(ii)`
From (i) and (ii), we have
`1-(x)/(2) = (2.94)/(4.9 xx 1.17)=0.5128`
or `(x)/(2) =1-0.5128=0.4872` or `x=0.9744` or `97.44%`
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Knowledge Check

  • The freezing point of a solution containing 0.2 g of acetic acid in 20.0 g benzene is lowered by 0.45^@C . The degree of association of acetic acid in benzene is (Assume acetic acid dimerises in benzene and K_f for benzene = 5.12 K kg "mol"^(-1) ) ?

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