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 ____

Precipitation Reaction Numerical – JEE Main PYQs | Mole Concept Chemistry

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

Step 1: Write the chemical reaction

NaI + AgNO₃ → AgI ↓ + NaNO₃

Step 2: Calculate molar mass of AgI

Molar mass of AgI = 108 + 127 = 235 g mol⁻¹

Step 3: Calculate moles of AgI formed

Moles of AgI = 4.74 / 235 = 0.0202 moles

Step 4: Use stoichiometric relation

From the equation, 1 mole of NaI produces 1 mole of AgI.
So, moles of NaI = 0.0202 moles

Step 5: Calculate molarity of NaI solution

Volume of NaI solution = 20 mL = 0.020 L
Molarity = moles / volume = 0.0202 / 0.020
≈ 1.01 M

Step 6: Nearest integer value

Molarity ≈ 1 M

✅ Final Answer: 1 M

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