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Chemistry Question 71 – JEE-MAIN 2025

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 mol1)

Determine the balanced chemical equation for the reaction between sodium iodide and silver nitrate.

Step 1: Write the Balanced Chemical Equation and Calculate Molar Mass of AgI✦ Active

The reaction between sodium iodide (NaI) and silver nitrate (AgNO3) produces silver iodide (AgI) precipitate and sodium nitrate (NaNO3). The balanced chemical equation is:

NaI(aq)+AgNO3(aq)AgI(s)+NaNO3(aq)

From the equation, 1 mole of NaI reacts to form 1 mole of AgI. Now, calculate the molar mass of AgI:

MAgI=MAg+MI=108+127=235 g/mol
Step 2: Calculate Moles of AgI and NaI○ Expand

Given the mass of AgI formed is 4.74 g. We can calculate the moles of AgI:

nAgI=mass of AgImolar mass of AgI=4.74 g235 g/mol0.02017 mol

Since the stoichiometric ratio between NaI and AgI is 1:1, the moles of NaI in the solution are equal to the moles of AgI formed:

nNaI=nAgI=0.02017 mol
💡 Teacher's Secret Hint

Ensure you use the correct stoichiometric ratio from the balanced equation.

Step 3: Calculate Molarity of NaI Solution○ Expand

The volume of the sodium iodide solution is 20 mL, which is 0.020 L. Molarity is defined as moles of solute per liter of solution:

Molarity of NaI=nNaIVolume of solution (L)=0.02017 mol0.020 L1.0085 M

Rounding to the nearest integer value, the molarity of the sodium iodide solution is 1 M.

💡 Teacher's Secret Hint

Remember to convert the volume from milliliters to liters before calculating molarity.

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