InterviewSolution
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InterviewSolution
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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)`. |
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Answer» Correct Answer - K=3.22 `"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` |
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