JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Some Basic Concepts of Chemistry

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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)

Answer: 1

Solution

Silver nitrate precipitates iodide quantitatively as silver iodide,

NaI + AgN O_{3} \to AgI + NaN O_{3}

, one mole of precipitate per mole of iodide. The molar mass of

AgI

108 + 127 = 235 g mol^{- 1}

, so the amount precipitated is

\frac{4.74}{235} = 0.0202 mol

, which equals the moles of sodium iodide taken. Dividing by the volume,

\frac{0.0202}{0.020 L}

, the molarity of the sodium iodide solution is

1 M

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Integer answer

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