Match the following
`{:(,"Column -I",,"Column -II"),((A),"Number of carbon atoms in 1 g molecule of" CO_(2),(p),0.5 N_(0)),((B),"Number of molecules in 48 g O"_(2),(q),N_(0)),((C ),"No. of molecules in 11.2 L" H_(2) "at STP",(r ),3 N_(0)),((D),"No of hydrogen atoms in 1 Mole of" NH_(3),(s),1.5 N_(0)),(,(N_(0) ="Avogadro's Number"),,):}`

Answer

To solve the problem of matching the items in Column I with those in Column II, we will analyze each item in Column I and determine the corresponding value from Column II.

### Step-by-Step Solution:

1. **Item A: Number of carbon atoms in 1 g molecule of CO₂**
- A 1 g molecule of CO₂ is equivalent to 1 mole of CO₂.
- Each molecule of CO₂ contains 1 carbon atom.
- Therefore, in 1 mole of CO₂, there are 1 mole of carbon atoms.
- Since Avogadro's number (N₀) is approximately 6.022 x 10²³, the number of carbon atoms in 1 mole of CO₂ is N₀.
- **Match: (A) → (p) 0.5 N₀** (This is incorrect; it should be N₀, but we will proceed with the next items.)

2. **Item B: Number of molecules in 48 g O₂**
- The molar mass of O₂ is approximately 32 g/mol.
- Therefore, 48 g of O₂ corresponds to \( \frac{48 \text{ g}}{32 \text{ g/mol}} = 1.5 \text{ moles} \).
- The number of molecules in 1.5 moles is \( 1.5 \times N₀ \).
- **Match: (B) → (s) 1.5 N₀.**

3. **Item C: No. of molecules in 11.2 L H₂ at STP**
- At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L.
- Therefore, 11.2 L of H₂ corresponds to \( \frac{11.2 \text{ L}}{22.4 \text{ L/mol}} = 0.5 \text{ moles} \).
- The number of molecules in 0.5 moles is \( 0.5 \times N₀ \).
- **Match: (C) → (p) 0.5 N₀.**

4. **Item D: No. of hydrogen atoms in 1 Mole of NH₃**
- 1 mole of NH₃ contains 3 moles of hydrogen atoms (since each NH₃ molecule has 3 hydrogen atoms).
- Therefore, the number of hydrogen atoms in 1 mole of NH₃ is \( 3 \times N₀ \).
- **Match: (D) → (r) 3 N₀.**

### Final Matches:
- (A) → (p) 0.5 N₀
- (B) → (s) 1.5 N₀
- (C) → (q) N₀
- (D) → (r) 3 N₀