Rearrange the following `(I ` to `IV)` in the order of increasing masses `:`
(I) `0.5` mole of `O_(3)" "(II) 0.5 gm` atom of oxygen
(III) `3.011 xx 10^(23)` molecules of `O_(2) " " (IV) 5.6` litre of `CO_(2)` at `STP`

Options

A. `II lt IV ltIII lt I`
B. `II lt Ilt IV ltIII`
C. `IV lt II lt III lt I`
D. `I lt II lt III lt IV`

Explanation

To solve the problem of rearranging the given substances in the order of increasing masses, we will calculate the mass of each substance step by step.

### Step 1: Calculate the mass of 0.5 mole of \( O_3 \)

1. **Molecular weight of \( O_3 \)**:
- Oxygen (O) has an atomic mass of approximately 16 g/mol.
- Therefore, the molecular weight of \( O_3 \) = \( 3 \times 16 = 48 \) g/mol.

2. **Mass calculation**:
- Using the formula:
\[ \text{Mass} = \text{moles} \times \text{molecular weight} \]
- For \( 0.5 \) mole of \( O_3 \):
\[ \text{Mass} = 0.5 \times 48 = 24 \text{ grams} \]

### Step 2: Calculate the mass of 0.5 gram atom of oxygen

1. **Molecular weight of oxygen**:
- The atomic weight of oxygen is 16 g/mol.
- Therefore, the mass of \( 0.5 \) gram atom of oxygen:
\[ \text{Mass} = 0.5 \text{ gram atom} \times 16 \text{ g/mol} = 8 \text{ grams} \]

### Step 3: Calculate the mass of \( 3.011 \times 10^{23} \) molecules of \( O_2 \)

1. **Number of moles**:
- Using Avogadro's number \( N_A = 6.022 \times 10^{23} \):
\[ \text{Moles} = \frac{3.011 \times 10^{23}}{6.022 \times 10^{23}} \approx 0.5 \text{ moles} \]

2. **Molecular weight of \( O_2 \)**:
- The molecular weight of \( O_2 \) = \( 2 \times 16 = 32 \) g/mol.

3. **Mass calculation**:
- For \( 0.5 \) moles of \( O_2 \):
\[ \text{Mass} = 0.5 \times 32 = 16 \text{ grams} \]

### Step 4: Calculate the mass of 5.6 liters of \( CO_2 \) at STP

1. **Molar volume at STP**:
- At STP, 1 mole of gas occupies 22.4 liters.
- Therefore, the number of moles in 5.6 liters:
\[ \text{Moles} = \frac{5.6 \text{ liters}}{22.4 \text{ liters/mole}} = 0.25 \text{ moles} \]

2. **Molecular weight of \( CO_2 \)**:
- The molecular weight of \( CO_2 \) = \( 12 + 2 \times 16 = 44 \) g/mol.

3. **Mass calculation**:
- For \( 0.25 \) moles of \( CO_2 \):
\[ \text{Mass} = 0.25 \times 44 = 11 \text{ grams} \]

### Summary of Masses Calculated:
- (I) \( O_3 \): 24 grams
- (II) 0.5 gram atom of oxygen: 8 grams
- (III) \( O_2 \): 16 grams
- (IV) \( CO_2 \): 11 grams

### Step 5: Arrange in order of increasing masses
- 8 grams (II) < 11 grams (IV) < 16 grams (III) < 24 grams (I)

### Final Order:
- (II) < (IV) < (III) < (I)