`{:("Column I (No. of moles)","Column II (Amount)"),("(A) 0.1 mole","(p) 4480 mL of CO"_(2)" at STP"),("(B) 0.2 mole","(q) 0.1 g atom of iron"),("(B) 0.25 mole","(r) "1.5xx10^(23)" molecules of oxygen gas"),("(C) 0.25 mole","(s) 9 mL of water"),("(D) 0.5 mole","(t) 200 mg of hydrogen gas"):}`

Answer

To solve the problem, we need to match the number of moles in Column I with the corresponding amounts in Column II. We will analyze each item in Column II and calculate the number of moles accordingly.

### Step-by-Step Solution:

1. **(A) 0.1 mole - (p) 4480 mL of CO₂ at STP**
- At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters (or 22400 mL).
- To find the number of moles in 4480 mL of CO₂:
\[ \text{Moles of CO₂} = \frac{\text{Volume (mL)}}{22400 \text{ mL/mole}} = \frac{4480 \text{ mL}}{22400 \text{ mL/mole}} = 0.2 \text{ moles} \]
- Therefore, **(p) corresponds to (B) 0.2 mole**.

2. **(B) 0.2 mole - (q) 0.1 g atom of iron**
- 1 gram atom of any element corresponds to 1 mole.
- Therefore, 0.1 gram atom of iron corresponds to:
\[ 0.1 \text{ g atom} = 0.1 \text{ moles} \]
- Thus, **(q) corresponds to (A) 0.1 mole**.

3. **(C) 0.25 mole - (r) 1.5 x 10²³ molecules of oxygen gas**
- To find the number of moles from the number of molecules, we use Avogadro's number (6.022 x 10²³ molecules/mole).
- The number of moles is calculated as:
\[ \text{Moles of O₂} = \frac{1.5 \times 10^{23} \text{ molecules}}{6.022 \times 10^{23} \text{ molecules/mole}} \approx 0.25 \text{ moles} \]
- Therefore, **(r) corresponds to (C) 0.25 mole**.

4. **(D) 0.25 mole - (s) 9 mL of water**
- The density of water is 1 g/mL, so 9 mL of water has a mass of:
\[ \text{Mass} = 9 \text{ g} \]
- The molar mass of water (H₂O) is approximately 18 g/mole.
- The number of moles is:
\[ \text{Moles of water} = \frac{\text{Mass (g)}}{\text{Molar mass (g/mole)}} = \frac{9 \text{ g}}{18 \text{ g/mole}} = 0.5 \text{ moles} \]
- Thus, **(s) corresponds to (D) 0.5 mole**.

5. **(E) 0.5 mole - (t) 200 mg of hydrogen gas**
- Convert 200 mg to grams:
\[ 200 \text{ mg} = 0.2 \text{ g} \]
- The molar mass of hydrogen gas (H₂) is approximately 2 g/mole.
- The number of moles is:
\[ \text{Moles of H₂} = \frac{0.2 \text{ g}}{2 \text{ g/mole}} = 0.1 \text{ moles} \]
- Therefore, **(t) corresponds to (E) 0.1 mole**.

### Final Matches:
- (A) 0.1 mole → (q) 0.1 g atom of iron
- (B) 0.2 mole → (p) 4480 mL of CO₂ at STP
- (C) 0.25 mole → (r) 1.5 x 10²³ molecules of oxygen gas
- (D) 0.5 mole → (s) 9 mL of water
- (E) 0.1 mole → (t) 200 mg of hydrogen gas