Question

Equivalent conductance of $ NaCl, HCl$ and $C_2H_5COONa $ at infinete dilution are $126.45, 426.16$ and $91\, \Omega ^{-1}\, cm^2 $ , respectively. The equivalent conductance of $C_2H_5COOH $ is

• A. $201.28 \Omega^{-1} \, cm^2 $
• B. $390.71 \Omega^{-1} \, cm^2 $
• C. $698.28 \Omega^{-1} \, cm^2 $
• D. $540.48 \Omega^{-1} \, cm^2 $

Answer 1

Answer: (B)

$\wedge_{C_{2} H_{5} COOH }=A_{C_{2} H _{5} COONa }+\wedge_{ HCl }-\wedge_{ NaCl }$
$=91.0+426.16-126.45=390.71\, ohm ^{-1}$