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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 )

Correct answer

0

Step-by-step solution

aligned & NaI _ ( aq ) + AgNO _ 3( aq) AgI _ ( s ) + NaNO ₃( aq ) & M , 20 ml excess & 4.74 ~g aligned Moles of I ⁻ in NaI = Moles of ( I ⁻ ) in AgI = 4.74 235 Moles of NaI = 4.74 235 Molarity [ NaI ]= 4.74 235 0.02 =1.008

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