During estimation of nitrogen by Dumas' method of compound X (0.42 g) :

_________ mL of $$N_2$$ gas will be liberated at STP. (nearest integer)
(Given molar mass in g mol$$^{-1}$$ : C : 12, H : 1, N : 14)

To determine the volume of $$N_2$$ gas liberated at STP using Dumas' method, we first analyze the molecular structure and composition of piperazine, $$C_4H_{10}N_2$$.

The molar mass of piperazine is calculated as follows:

• Carbon: $$4 \times 12 = 48\text{ g mol}^{-1}$$
• Hydrogen: $$10 \times 1 = 10\text{ g mol}^{-1}$$
• Nitrogen: $$2 \times 14 = 28\text{ g mol}^{-1}$$

Hence, the molar mass of piperazine is

$$48 + 10 + 28 = 86\text{ g mol}^{-1}.$$

For a sample mass of $$0.42\text{ g}$$, the number of moles of piperazine is

$$\frac{0.42}{86} \approx 0.0048837\text{ mol}.$$

Each molecule of piperazine contains two nitrogen atoms, which together correspond to one molecule of $$N_2$$. Therefore, one mole of piperazine liberates one mole of $$N_2$$ gas.

Hence,

$$\text{Moles of }N_2 = 0.0048837\text{ mol}.$$

If the molar volume of an ideal gas at STP is taken as $$22,400\text{ mL mol}^{-1}$$, then the volume of nitrogen liberated is

$$0.0048837 \times 22,400 \approx 109.39\text{ mL}.$$

Rounding to the nearest integer,

$$\boxed{109\text{ mL}}.$$

However, if the standard molar volume at STP is taken as $$22,700\text{ mL mol}^{-1}$$ (corresponding to STP defined at $$1$$ bar), then

$$0.0048837 \times 22,700 \approx 110.86\text{ mL},$$

which rounds to

$$\boxed{111\text{ mL}}.$$

Thus, the slight difference in the final answer arises solely from the convention adopted for the molar volume of an ideal gas at STP. Using $$22.4\text{ L mol}^{-1}$$ gives $$109\text{ mL}$$, whereas using $$22.7\text{ L mol}^{-1}$$ gives $$111\text{ mL}$$.