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Conductivity of $2\times 10^{^{-3}}M$ methanoic acid is $8\times 10^{-5}S\; cm^{-1}$ . Calculate its molar conductivity and degree of dissociation if $\wedge ^{\circ}m$ for methanoic acid is $404\; S\; cm^{2}\; mol^{-1}$

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$\Lambda$ _m = k x 1000/ M = 8 x 10^-5 x 1000 / 2 x 10^-3 = 40 Scm2mol-1

α= $\Lambda$ _m / $\Lambda$ _m^o= 40/ 404 = 0.099

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