$20 \ mL$ of sodium iodide solution gave $4.74 \ g$ silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is $........ M$. $(Nearest \ Integer \ value)$
$(Given: Na=23, I=127, Ag=108, N=14, O=16 \ g \ mol^{-1})$

  • A
    $0$
  • B
    $2$
  • C
    $1$
  • D
    $9$

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