The question as shown is incomplete because the identity/formula of compound $X$ is missing. In Dumas' method, the volume of $\mathrm{N_2}$ liberated depends on the number of nitrogen atoms per molecule and the molar mass of $X$.

1. General relation

If compound $X$ has molecular formula with:

• molar mass $M$
• $n$ nitrogen atoms per molecule

then moles of compound in $0.42\,\text{g}$ are $\frac{0.42}{M}.$

Since each molecule containing $n$ nitrogen atoms gives $\frac{n}{2}$ moles of $\mathrm{N_2}$,

moles of $\mathrm{N_2}$ formed are $\frac{0.42}{M}\cdot \frac{n}{2}.$

At STP, volume is $V=\frac{0.42}{M}\cdot \frac{n}{2}\cdot 22400\,\text{mL}.$ So, $V=\frac{4704\,n}{M}\,\text{mL}.$

2. Using the stored answer

Stored answer is $109\,\text{mL}$. Hence $\frac{4704n}{M}\approx 109.$ So, $M\approx \frac{4704n}{109}.$

For common organic nitrogen compounds, if $n=1$, $M\approx \frac{4704}{109}\approx 43.2,$ which is not a typical integer molar mass for a stable simple CHN compound matching usual exam patterns.

If $n=2$, $M\approx \frac{9408}{109}\approx 86.3,$ which suggests a compound of molar mass about $86$ containing two nitrogen atoms. Then the exact volume would be $V=\frac{4704\times 2}{86}=109.4\,\text{mL},$ whose nearest integer is $109\,\text{mL}.$

Thus the stored answer is consistent with a missing compound formula having 2 nitrogen atoms and molar mass $86$.

3. Conclusion

Because the formula of compound $X$ is absent in the provided question image/text, the answer cannot be derived uniquely from the visible data alone.

However, the stored answer $109$ mL is internally consistent with a likely omitted formula for $X$.