A sample of urine containing `0.3 g` of urea was treated with an excess of `0.2 M` nitrous acid, according to the equation.
`NH_(2)CONH_(2) + 2HNO_(2) rarr CO_(2) + 2N_(2) + 2H_(2)O`
The gass produced passed through aqueous `KOH` solution and the final valume is measured.
(Given, `Mw_("urea") = 60 g mol^(-1)`, molar volume of gas at standard condition, i.e., at room temperature `25^(@)C` and 1 atm pressure. `RTP` (room temperature pressure) also is `24.4 L` or `24400 mL mol^(-1))`
What is the volume at `RTP`?

Options

A. `122 mL`
B. `244 mL`
C. `366 mL`
D. `488 mL`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the moles of urea
Given:
- Mass of urea = 0.3 g
- Molar mass of urea = 60 g/mol

Using the formula:
\[ \text{Moles of urea} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.3 \, \text{g}}{60 \, \text{g/mol}} = 0.005 \, \text{mol} \, (5 \, \text{mmol}) \]

### Step 2: Determine the moles of gases produced
From the balanced chemical equation:
\[ \text{NH}_2\text{CONH}_2 + 2\text{HNO}_2 \rightarrow \text{CO}_2 + 2\text{N}_2 + 2\text{H}_2\text{O} \]
1 mole of urea produces:
- 1 mole of CO₂
- 2 moles of N₂

Thus, for 0.005 moles of urea:
- Moles of CO₂ produced = 0.005 mol
- Moles of N₂ produced = \(2 \times 0.005 \, \text{mol} = 0.010 \, \text{mol}\)

### Step 3: Understand the effect of KOH
The KOH solution absorbs CO₂, so only N₂ gas will be measured. Therefore, we will only consider the moles of N₂ gas produced:
- Moles of N₂ = 0.010 mol

### Step 4: Calculate the volume of N₂ at RTP
At room temperature and pressure (RTP), 1 mole of gas occupies 24.4 L (or 24400 mL).

Using the formula:
\[ \text{Volume} = \text{moles} \times \text{molar volume} \]
\[ \text{Volume of N₂} = 0.010 \, \text{mol} \times 24400 \, \text{mL/mol} = 244 \, \text{mL} \]

### Final Answer
The volume of nitrogen gas at RTP is **244 mL**.

---