0.30 gm of an organic compound gave 50 ml of nitrogen collected at 300 K and 715 mm pressure in dumas method. Calculate the percentage of nitrogen in the compound. (Vapour pressure of water or aqueous tension of water at 300 K is 15 mm.)

Options

A. 0.2238 *100
B. 0.1746*100
C. 0.5511*100
D. 0.8274*100

Explanation

To solve the problem of calculating the percentage of nitrogen in the organic compound, we will follow these steps:

### Step 1: Calculate the pressure of nitrogen gas (P1)
The total pressure (P_total) is given as 715 mm, and the vapor pressure of water at 300 K is 15 mm. Therefore, the pressure of nitrogen gas (P1) is calculated as follows:

\[ P1 = P_{total} - P_{water} = 715 \, \text{mm} - 15 \, \text{mm} = 700 \, \text{mm} \]

### Step 2: Convert the volume of nitrogen gas (V1) to liters
The volume of nitrogen gas (V1) is given as 50 ml. We convert this to liters:

\[ V1 = 50 \, \text{ml} = 0.050 \, \text{L} \]

### Step 3: Use the ideal gas law to find the volume of nitrogen at STP (V2)
We will use the formula relating the volumes, pressures, and temperatures of the gas:

\[ \frac{P1 \cdot V1}{T1} = \frac{P2 \cdot V2}{T2} \]

Where:
- \(P1 = 700 \, \text{mm}\)
- \(V1 = 0.050 \, \text{L}\)
- \(T1 = 300 \, \text{K}\)
- \(P2 = 760 \, \text{mm}\) (standard pressure)
- \(T2 = 273 \, \text{K}\) (standard temperature)

Rearranging the equation to solve for \(V2\):

\[ V2 = \frac{P1 \cdot V1 \cdot T2}{T1 \cdot P2} \]

Substituting the values:

\[ V2 = \frac{700 \, \text{mm} \cdot 0.050 \, \text{L} \cdot 273 \, \text{K}}{300 \, \text{K} \cdot 760 \, \text{mm}} \]

Calculating \(V2\):

\[ V2 = \frac{700 \cdot 0.050 \cdot 273}{300 \cdot 760} \approx 0.0419 \, \text{L} \approx 41.9 \, \text{ml} \]

### Step 4: Calculate the number of moles of nitrogen (N2)
Using the molar volume of a gas at STP (22.4 L/mol):

\[ \text{Number of moles of } N2 = \frac{V2}{22.4 \, \text{L/mol}} = \frac{0.0419 \, \text{L}}{22.4 \, \text{L/mol}} \approx 0.00187 \, \text{mol} \]

### Step 5: Calculate the mass of nitrogen in the compound
The molecular weight of nitrogen (N2) is approximately 28 g/mol. Therefore, the mass of nitrogen in the compound is:

\[ \text{Mass of } N2 = \text{Number of moles} \times \text{Molar mass} = 0.00187 \, \text{mol} \times 28 \, \text{g/mol} \approx 0.052 \, \text{g} \]

### Step 6: Calculate the percentage of nitrogen in the compound
The percentage of nitrogen in the organic compound is calculated as follows:

\[ \text{Percentage of } N2 = \left( \frac{\text{Mass of } N2}{\text{Mass of compound}} \right) \times 100 = \left( \frac{0.052 \, \text{g}}{0.30 \, \text{g}} \right) \times 100 \approx 17.45\% \]

### Final Answer
The percentage of nitrogen in the compound is approximately **17.45%**.

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