JEE Chemistry 2025 Question with Solution

Given: Volume of sodium iodide solution = $20 \, \text{mL}$, mass of silver iodide formed = $4.74 \, \text{g}$.

Find: The molarity of the sodium iodide solution.

The reaction is

$$\text{NaI} + \text{AgNO}_3 \rightarrow \text{AgI} + \text{NaNO}_3$$

Molar mass of $\text{AgI}$ is

$$108 + 127 = 235 \, \text{g mol}^{-1}$$

Moles of $\text{AgI}$ formed are

$$\frac{4.74}{235} = 0.02017 \, \text{mol}$$

From the balanced equation, $\text{NaI}$ and $\text{AgI}$ are in $1:1$ ratio, so moles of $\text{NaI}$ = $0.02017 \, \text{mol}$.

Volume of solution in litres is

$$20 \, \text{mL} = 0.020 \, \text{L}$$

Therefore, molarity is

$$M = \frac{0.02017}{0.020} = 1.0085 \, \text{M}$$

Nearest integer value is $1$.

Therefore, the molarity of the sodium iodide solution is $1 \, \text{M}$.

Given: Silver iodide precipitate mass = $4.74 \, \text{g}$, sodium iodide solution volume = $20 \, \text{mL}$.

Find: Molarity of sodium iodide solution.

1. Write the balanced chemical equation:

$$\text{NaI} + \text{AgNO}_3 \rightarrow \text{AgI} \downarrow + \text{NaNO}_3$$

2. Calculate molar mass of $\text{AgI}$:

$$108 + 127 = 235 \, \text{g mol}^{-1}$$

3. Calculate moles of $\text{AgI}$ precipitated:

$$\text{Moles of AgI} = \frac{4.74 \, \text{g}}{235 \, \text{g mol}^{-1}} = 0.0202 \, \text{mol}$$

4. Use stoichiometry:

$$\text{Moles of NaI} = \text{Moles of AgI} = 0.0202 \, \text{mol}$$

5. Convert volume into litres:

$$20 \, \text{mL} = 0.020 \, \text{L}$$

6. Apply molarity formula:

$$\text{Molarity} = \frac{\text{Moles}}{\text{Volume in L}} = \frac{0.0202}{0.020} = 1.01 \, \text{M}$$

After rounding to the nearest integer, the answer is $1$.

The numerical value answer is 1.