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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) ] |
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Answer» Solution :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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