Question

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$^{-1}$)

Hint:

- In precipitation reactions, stoichiometry is 1:1 for simple salts - Always convert mass to moles using molar mass - Remember to convert mL to L for molarity calculations - Nearest integer rounding for final answer

Answer 1

Correct Answer: 1

Step 1: Write the balanced chemical equation \[ \text{NaI} + \text{AgNO}_3 \rightarrow \text{AgI} \downarrow + \text{NaNO}_3 \]
Step 2: Calculate moles of AgI precipitated Molar mass of AgI = \( 108 + 127 = 235 \) g mol\(^{-1}\) \[ \text{Moles of AgI} = \frac{4.74 \text{ g}}{235 \text{ g mol}^{-1}} = 0.0202 \text{ mol} \]
Step 3: Determine moles of NaI From stoichiometry, 1 mole NaI produces 1 mole AgI \[ \text{Moles of NaI} = 0.0202 \text{ mol} \]
Step 4: Calculate molarity of NaI solution \[ \text{Molarity} = \frac{\text{Moles}}{\text{Volume in L}} = \frac{0.0202 \text{ mol}}{0.020 \text{ L}} = 1.01 \text{ M} \] Rounding to nearest integer: \[ \text{Molarity} \approx 1 \text{ M} \]