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).

 

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).

📷 Question Image:

---

Short Solution (Text):

Step 1: Write reaction

$$NaI+AgN{O}_3\to AgI\downarrow +NaN{O}_{\mathrm{3}}$$

1 mole NaI gives 1 mole AgI.

---

Step 2: Moles of AgI formed

Molar mass of AgI = 107.87 (Ag) + 126.90 (I) = 234.77 g/mol

$$\text{Moles of AgI}=\frac{4.74}{234.77}\approx 0.0202\text{ mol}$$

---

Step 3: Moles of NaI in solution

Since 1:1 ratio, moles of NaI = moles of AgI = 0.0202 mol

---

Step 4: Molarity of NaI solution

Volume of solution = 20 mL = 0.020 L

$$M=\frac{\text{moles}}{\text{volume (L)}}=\frac{0.0202}{0.020}\approx 1.01M$$

---

✅ Final Answer:
The molarity of sodium iodide solution = 1 M (nearest integer).

---

📷 Solution Image:

Conclusion – Video Solution:

By using the simple 1:1 reaction between NaI and AgNO₃, calculating moles of AgI formed, and dividing by the volume of solution, we get the molarity as ≈ 1 M.