Conductivity of 0.001M acetic acid at a certain temperature is 5.07×10^-5 S/cm.if limiting molar conductivity of acetic acid at the same temperature is 390 S cm^2/mol.the dissociation constant of acetic acid at that temperature is?

We're at a warmer temperature than #25^@ "C"#.

We know that the limiting molar conductivity, #Lambda_m^0#, is simply the molar conductivity as if the electrolyte ionized completely. #Lambda_m# is the value measured normally...

Thus, it is weighted by the fraction of dissociation #alpha# (Eq. 7) to give the actual molar conductivity #Lambda_m# at the given concentration:

#Lambda_m = alphaLambda_m^0#

The conductivity you gave is not in the proper units yet.

#5.07 xx 10^(-5) "S/""cm" xx cancel"1 L"/("0.001 mols") xx (1000 cancel"mL")/cancel"1 L" xx ("1 cm"^3)/cancel"1 mL"#

#= "50.7 S"cdot"cm"^2"/mol"#

And now this can be compared to #Lambda_m^0# to find the fraction of dissociation:

#alpha = Lambda_m/Lambda_m^0#

#= ("50.7 S"cdot"cm"^2"/mol")/("390 S"cdot"cm"^2"/mol")#

#= 0.13#

By definition, the dissociation of acetic acid is given by

#"CH"_3"COOH"(aq) rightleftharpoons "CH"_3"COO"^(-)(aq) + "H"^(+)(aq)#

with mass action expression

#K_a = (["CH"_3"COO"^(-)]["H"^(+)])/(["CH"_3"COOH"])#

Some fraction #alpha# of the starting concentration becomes the acetate and proton products, and that much is lost RELATIVE to #100%# of the starting acid, so...

#K_a = ((alphacdot["CH"_3"COOH"]_0)(alphacdot["CH"_3"COOH"]_0))/((1 - alpha)cdot["CH"_3"COOH"]_0)#

#= (alpha^2["CH"_3"COOH"]_0)/(1 - alpha)# (cf. Eq. 21)

#= (0.13^2 cdot "0.001 M")/(1 - 0.13)#

And we obtain an acid dissociation constant of:

#color(blue)(K_a = 1.94 xx 10^(-5))#

at whatever temperature this is.

The actual value is around #1.76 xx 10^(-5)# at #25^@ "C"#, and dissociation is endothermic, so a higher #K_a# means this temperature is warmer than #25^@ "C"#.