Question

At 715 mm pressure, 300 K, volume of N\(_2\) (g) evolved was 80 mL by a 0.4 g sample of organic compound. Find the percentage of N in the organic compound. Given aqueous tension at 300 K = 15 mm.

• A. 20.95
• B. 25.85
• C. 30.25
• D. 15.83

Hint:

In gas-related problems, remember to subtract the aqueous tension from the total pressure before applying the Ideal Gas Law.

Answer 1

The Correct Option is D

We are given the following data:
- Pressure \(P = 715 \, \text{mm Hg}\)
- Volume of evolved gas \(V = 80 \, \text{mL}\)
- Temperature \(T = 300 \, \text{K}\)
- Mass of organic compound \(m = 0.4 \, \text{g}\)
- Aqueous tension at 300 K = 15 mm Hg We need to calculate the percentage of nitrogen in the organic compound. 1. First, we use the Ideal Gas Law to find the number of moles of N\(_2\): \[ PV = nRT \] Where: - \(P = 715 - 15 = 700 \, \text{mm Hg}\) (subtract aqueous tension), - \(R = 0.0821 \, \text{L atm} / \text{mol K}\),
- \(T = 300 \, \text{K}\),
- \(V = 80 \, \text{mL} = 0.08 \, \text{L}\).
Rearranging the Ideal Gas Law to solve for \(n\) (moles of N\(_2\)): \[ n = \frac{PV}{RT} \] Substituting the values: \[ n = \frac{700 \times 0.08}{0.0821 \times 300} \] \[ n = 0.0284 \, \text{mol} \] 2. Now, we calculate the mass of nitrogen in the sample: - The molar mass of N\(_2\) is \(28 \, \text{g/mol}\). - The mass of nitrogen is: \[ \text{Mass of N} = n \times \text{Molar mass of N}_2 = 0.0284 \times 28 = 0.7952 \, \text{g} \] 3. Finally, we calculate the percentage of nitrogen in the organic compound: \[ \% \, \text{of N} = \frac{\text{Mass of N}}{\text{Mass of sample}} \times 100 = \frac{0.7952}{0.4} \times 100 = 15.83\% \] Thus, the percentage of nitrogen in the organic compound is \(15.83\%\).