JEE Main 2025 Apr 8 Shift 2, Chemistry Q23: Mole Concept And Equivalent Concept

JEE Main2025Apr 8, Shift 2Chemistry

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 $^{- 1}$ )

Your answer

Solution

The reaction is $N a I (a q) + A g N O_{3} (a q) \to A g I (s) + N a N O_{3} (a q)$ , so the moles of I $^{-}$ originally present in NaI equal the moles of AgI precipitate formed.

Molar mass of AgI $= 108 + 127 = 235$ g/mol, so moles of AgI $= \frac{4.74}{235} \approx 0.02017$ mol.

This equals the moles of NaI present in the original 20 mL sample.

Molarity $= \frac{moles}{volume in litres} = \frac{0.02017}{0.020} \approx 1.008$ M.

Rounded to the nearest integer, the molarity is 1 M.

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JEE Main 2025 Apr 8 Shift 2, Chemistry Q23: Mole Concept And Equivalent Concept

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