112% labelled oleum is diluted with sufficient water. The solution on mixing with `5.3 gm Na_(2)CO_(3)` liberates `CO_(2)`. The volume of `Co_(2)` given out at 1 atm at 273 K will be

Options

A. 1.12 litres
B. 1.23 lit
C. 2.2 lit
D. 37.75 lit

Answer

To solve the problem, we need to follow these steps:

### Step 1: Write the reaction
When sodium carbonate (Na₂CO₃) reacts with oleum (which contains sulfuric acid, H₂SO₄), carbon dioxide (CO₂) is liberated. The balanced chemical equation for this reaction is:

\[ \text{Na}_2\text{CO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} + \text{CO}_2 \]

### Step 2: Calculate moles of Na₂CO₃
To find the moles of Na₂CO₃, we use the formula:

\[ \text{Moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

The molar mass of Na₂CO₃ is:

- Na: 23 g/mol (2 Na = 46 g)
- C: 12 g/mol
- O: 16 g/mol (3 O = 48 g)

So, the molar mass of Na₂CO₃ = 46 + 12 + 48 = 106 g/mol.

Now, using the mass given (5.3 g):

\[ \text{Moles of Na}_2\text{CO}_3 = \frac{5.3 \, \text{g}}{106 \, \text{g/mol}} \approx 0.050 \, \text{mol} \]

### Step 3: Determine moles of CO₂ produced
From the balanced equation, we see that 1 mole of Na₂CO₃ produces 1 mole of CO₂. Therefore, the moles of CO₂ produced will also be:

\[ \text{Moles of CO}_2 = 0.050 \, \text{mol} \]

### Step 4: Calculate the volume of CO₂ at STP
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Thus, the volume of CO₂ produced can be calculated as:

\[ \text{Volume of CO}_2 = \text{Moles of CO}_2 \times 22.4 \, \text{L/mol} \]

Substituting the moles of CO₂:

\[ \text{Volume of CO}_2 = 0.050 \, \text{mol} \times 22.4 \, \text{L/mol} \approx 1.12 \, \text{L} \]

### Final Answer
The volume of CO₂ liberated at 1 atm and 273 K is approximately **1.12 liters**.

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