NEET Questions Solved


Equivalent conductances of NaCl, HCl and C2H5COONa at infinite dilution are 126.45, 426.16 and 91Ω-1cm2. The equivalent conductance of C2H5COOH is

(a) 201.28 Ω-1cm2

(b) 390.71 Ω-1cm2

(c) 698.28 Ω-1cm2

(d) 540.48 Ω-1cm2

(b) By Kohlrausch's law 

                λ for NaCl = λNa+ + λCl-               ...(i)                 λ for HCl = λH+ + λCl-                ...(ii)λ for C2H5COONa = λNa+ + λC2H5COO-   ...(iii)So, λ for C2H5COOHOn adding Eqs. (ii) and (iii) and then subtractingEq. (i)             =λ of C2H5COONa + λ of HCl - λ for NaCl             =(91 + 426.16 - 126.45)Ω-1cm2             = 390.71 Ω-1cm2

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