The freezing point of `0.02 mol` fraction solution of acetic acid (A) in benzene (B) is `277.4K`. Acetic acid exists partly as a dimer `2A = A_(2)`. Calculate equilibrium constant for the dimerisation. Freezing point of benzene is `278.4K` and its heat of fusion `DeltaH_(f)` is `10.042 kJ mol^(-1)`.

Answer

K=3.22

Explanation

`"X"_("B")=0.02implies"n"_("B")=0.02" and"" n"_("A")=0.98`
`"W"_("A")=0.98xx78"gm"`
Molality(m) `=("n"_("B")xx1000)/"W"_("A")=(0.02xx1000)/(0.98xx78)=0.2616`
`"K"_("f")=("RT"_("f")^(2)"m"_("A"))/(Delta_("f")"H"xx1000)=(8.314xx(278.4)^(2)78)/(10042xx1000)`
`=5.00"K"//"m"`
`Delta"T"_("F")="i"xx"K"_("F")"m"`
`1="i"xx5xx0.2616`
`"i"=1/(5xx0.2616)`
`=0.7645`
`"i"=1+(1/"n"-1)alpha`
As acetic acid dimerise, n=2
`"i"=1-alpha/2`

`0.7645=1-alpha/2`
`alpha=0.4710`
We assume: Molarity=Molality
`therefore "Molarity (C)"=0.2616`
`2"A "hArr "A"_(2)`
`{:("At" t = 0,C,0),("At equi.",C(1-a),(Calpha)/(2)):}`
`"K"=(["A"_(2)])/(["A"]^(2))=("C"alpha)/(2"C"^(2)(1-alpha)^(2))=alpha/(2"C"(1-alpha)^(2))=0.471/(2xx0.2616(0.529)^(2))=3.22`