The number of ions present in $2.0 \ L$ of a solution of $0.8 \ M \ K_4[Fe(CN)_6]$ is

  • A
    $4.8 \times 10^{24}$
  • B
    $4.8 \times 10^{23}$
  • C
    $9.6 \times 10^{24}$
  • D
    $9.6 \times 10^{23}$

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