20.27g of benzene containing 0.2965g of benzoic acid (mol.wt. =122)freezes at 0.137^(@) below the freezing point of pure benzene. If benzoic acid exists as dimer in benzene, find its degree of association. K_(f) for benzene is 5.12^(@)C.m^(-1)

20.27g of benzene containing 0.2965g of benzoic acid (mol.wt. =122)fre

1.	20.27g of benzene containing 0.2965g of benzoic acid (mol.wt. =122)freezes at 0.137^(@) below the freezing point of pure benzene. If benzoic acid exists as dimer in benzene, find its degree of association. K_(f) for benzene is 5.12^(@)C.m^(-1)
Answer» Solution :Molality (calculated) `=(0.2965//122)/(20.27)xx1000` `=0.12` mole//1000g Molality (observed) `(DeltaT_(f))/(K_(f))=(0.317)/(5.12)=0.0619` mole/1000g `:.i=("molality (observed)")/("molality(calculated)")=(0.0619)/(0.12)=0.5158`……(Eqn 10) Since benzoic acid exists as dimer in BENZENE we have `{:("moles BEFOR association",,1"mole",,,,0),(,,2C_(6)H_(5)COOH,,=,,(C_(6)H_(5)COOH)_(2)),("moles after association",,(1-x),,,,(x)/(2)(x="degree of association")):}` `:.i=(1-x+(x)/(2))/(1)=0.5158` ........(Eqn. 10) `x=0.9684` or `96.84%`