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The freezing point of benzene decreases by 0.45^(@)C when 0.2 g of acetic acidis added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be ……. (K_(f) for benzene = 5.12 K kg mol^(-1))

Answer»

`64.6%`
`80.4%`
`74.6%`
`94.6%`

Solution :`Delta T_(f)=0.45`
`m=(((0.2)/(60))xx1000)/(20)=(1)/(6)`
`K_(f)=5.12` K kg / mol
`i=1+((1)/(n)-1)x "" (n=2)`
`= 1-(x)/(2)`
`Now, Delta T_(f)=iK_(f)m`
`0.45=(1-(x)/(2))(5.12)((1)/(6))`
x = 0.94
`THEREFORE` PERCENTAGE of `~~ 94%`


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