A yellow solid X has a molar mass of 138 g mol⁻¹…

Step 1. Dry nitrogen at STP. The gas is wet, so $p_1 = 750 - 27 = 723\ \text{mm Hg}$. From $\dfrac{p_1V_1}{T_1} = \dfrac{p_0V_0}{T_0}$ we want $V_0$, so $V_0 = \dfrac{p_1V_1T_0}{p_0T_1}$. Then $V_0 = 51.8\ \text{mL} \times \dfrac{723}{760} \times \dfrac{273}{300}$ $= 44.84\ \text{mL}$.

Step 2. Dumas percentage. Mass of N $= \dfrac{28\ \text{g}}{22400\ \text{mL}} \times 44.84\ \text{mL} = 0.0561\ \text{g}$. So $\%\ \text{N} = \dfrac{0.0561\ \text{g}}{0.276\ \text{g}} \times 100 = 20.31\%$.

Step 3. Kjeldahl ammonia. Acid taken: $50.0\ \text{mL} \times 0.100\ \text{mol L}^{-1} = 5.00\ \text{mmol}$. Alkali used: $30.0\ \text{mL} \times 0.200\ \text{mol L}^{-1} = 6.00\ \text{mmol}$, which neutralises $3.00\ \text{mmol}$ of acid. The ammonia took $5.00 - 3.00 = 2.00\ \text{mmol}$ of acid, so it was $2 \times 2.00 = 4.00\ \text{mmol}$.

Step 4. Kjeldahl percentage. Mass of N $= 4.00\ \text{mmol} \times 14\ \text{mg mmol}^{-1} = 0.056\ \text{g}$. So $\%\ \text{N} = \dfrac{0.056\ \text{g}}{0.552\ \text{g}} \times 100 = 10.14\%$.

Step 5. Atoms per molecule. One mole, $138\ \text{g}$, holds $28.03\ \text{g}$ of N by Dumas: $\dfrac{28.03}{14} = 2.00$, so $2$ atoms. Kjeldahl finds only $14.00\ \text{g}$: $\dfrac{14.00}{14} = 1.00$, so $1$ atom.

Why the methods disagree. Dumas' method sets free all the nitrogen as $\text{N}_2$. Kjeldahl's method does not work for nitrogen in nitro and azo groups, or in a ring as in pyridine, because that nitrogen does not change to ammonium sulphate. So one of the two nitrogen atoms of X sits in such a group. X could be a nitroaniline, $\text{C}_6\text{H}_6\text{N}_2\text{O}_2$, whose amino nitrogen is found and whose nitro nitrogen is missed.