Addition of excess of $AgNO_3$ to an aqueous solution of $1$ mole of $PdCl_2 \cdot 4 NH_3$ gives $2$ moles of $AgCl$. The conductivity of this solution corresponds to

  • A
    $1:2$ electrolyte
  • B
    $1:4$ electrolyte
  • C
    $1:1$ electrolyte
  • D
    $1:3$ electrolyte

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