Equivalent conductance of an electrolyte containing $\mathrm{NaF}$ at infinite dilution is $90.1$ $\mathrm{Ohm}^{-1} \mathrm{~cm}^{2}$. If $\mathrm{NaF}$ is replaced by $\mathrm{KF}$ what is the value of equivalent conductance?

• Correct answer
  A

  $90.1 \mathrm{Ohm}^{-1} \mathrm{~cm}^{2}$
• B

  $111.2 \mathrm{Ohm}^{-1} \mathrm{~cm}^{2}$
• C

  0
• D

  $222.4 \mathrm{Ohm}^{-1} \mathrm{~cm}^{2}$

Hence, the correct answer is option: 'A'.