1. Given data

• Mass of organic compound $= 1.0\,\text{g}$
• Volume of $N_2 = 150\,\text{mL} = 0.150\,\text{L}$
• Temperature $= 300\,\text{K}$
• Total pressure $= 900\,\text{mm Hg}$
• Aqueous tension at $300\,\text{K} = 15\,\text{mm Hg}$

2. Correct pressure of dry nitrogen

Since nitrogen is collected over water, $P_{N_2} = 900 - 15 = 885\,\text{mm Hg}$

Convert into atm: $P_{N_2} = \frac{885}{760}\,\text{atm}$

3. Use ideal gas equation

$PV = nRT$

So, $n = \frac{PV}{RT}$

Substituting values: $n = \frac{\left(\frac{885}{760}\right)(0.150)}{(0.0821)(300)}$

Now calculate: $\frac{885}{760} \approx 1.1645$

$n \approx \frac{1.1645 \times 0.150}{24.63}$

$n \approx \frac{0.1747}{24.63} \approx 0.00709\,\text{mol of } N_2$

4. Mass of nitrogen present

1 mole of $N_2$ has mass $28\,\text{g}$.

Therefore, $\text{mass of nitrogen} = 0.00709 \times 28$

$\approx 0.1985\,\text{g}$

5. Percentage of nitrogen

$\%N = \frac{0.1985}{1.0} \times 100$

$\%N \approx 19.85\%$

Nearest integer: $\boxed{20}$

6. Comparison with stored answer

Stored correct answer = $20$

Our derived answer is also $20$.