Calculate Molarity from Mass of Precipitate

❓ Question

20 mL of sodium iodide solution gave:

$$4.74\text{ g}$$

silver iodide when treated with excess silver nitrate solution.

The molarity of sodium iodide solution is:

$$\_\_\_\_\_ \text{ M}$$

(Nearest integer value)

Given:

$$Na=23,\ I=127,\ Ag=108$$

---

🖼 Question Image

---

✍️ Short Explanation

This problem is based on:

👉 Mole concept
👉 Precipitation reaction
👉 Molarity calculation.

Main idea:

$$NaI+AgNO_3\rightarrow AgI\downarrow +NaNO_3$$

Mole ratio:

$$\boxed{ NaI:AgI=1:1 }$$

---

🔷 Step 1 — Calculate Molar Mass of AgI 💯

$$M(AgI)=108+127$$
$$=235\text{ g/mol}$$

---

🔷 Step 2 — Find Moles of AgI

Given mass:

$$4.74\text{ g}$$

So:

$$\text{Moles of AgI} = \frac{4.74}{235}$$
$$\approx0.0202$$

---

🔷 Step 3 — Use Stoichiometry

Reaction:

$$NaI+AgNO_3\rightarrow AgI+NaNO_3$$

Mole ratio:

$$1:1$$

Thus:

$$\text{Moles of NaI}=0.0202$$

---

🔷 Step 4 — Convert Volume into Litres

Given volume:

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

---

🔷 Step 5 — Calculate Molarity

Formula:

$$\boxed{ M=\frac{\text{moles}}{\text{volume in litres}} }$$
$$M=\frac{0.0202}{0.020}$$
$$\approx1.01$$

Nearest integer:

$$\boxed{1}$$

---

🔷 Step 6 — JEE Trap Alert 🚨

❌ mL ko litre me convert na karna

❌ AgI ka molar mass galat lena

❌ Mole ratio ignore kar dena

Remember:

$$\boxed{ 1\text{ mole}= \frac{\text{mass}}{\text{molar mass}} }$$

---

✅ Final Answer

$$\boxed{1\text{ M}}$$

---

📚 Related Topics

• MOI of Combined Bodies Using Parallel Axis
• Van’t Hoff Factor Shortcut for Percentage Dissociation
• Quadratic Equation Roots Reciprocal Relation Trick