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NVAEasyJEE 2025Quantitative Analysis (C, H, N…)

JEE Chemistry 2025 Question with Solution

During estimation of Nitrogen by Dumas' method of compound X (0.42 g0.42 \, \text{g}) :

Cyclic six-membered ring structure of compound X with two nitrogen atoms opposite each other and one hydrogen attached to each nitrogen.

mL of N2N_2 gas will be liberated at STP. (nearest integer) \text{(Given molar mass in g mol^{-1} : C : 12, H : 1, N : 14)}

Answer

Correct answer:109

Step-by-step solution

Standard Method

Given: Compound XX has the shown structure and mass 0.42 g0.42 \, \text{g}.

Find: Volume of N2N_2 liberated at STP.

The structure corresponds to piperazine with molecular formula C4H10N2C_4H_{10}N_2.

Molecular weight calculation:

4×12+10×1+2×14=48+10+28=86 g/mol4 \times 12 + 10 \times 1 + 2 \times 14 = 48 + 10 + 28 = 86 \, \text{g/mol}

Moles of compound:

0.4286≈0.0048845  mol\frac{0.42}{86} \approx 0.0048845 \; \text{mol}

In Dumas' method, all nitrogen of the compound is converted into N2N_2. Since C4H10N2C_4H_{10}N_2 contains two nitrogen atoms, 11 mole of compound gives 11 mole of N2N_2.

So, moles of N2N_2 produced are

0.0048845  mol0.0048845 \; \text{mol}

At STP, volume of gas is

0.0048845×22.4=0.109  L0.0048845 \times 22.4 = 0.109 \; \text{L}

Converting to mL,

0.109×1000=109  mL0.109 \times 1000 = 109 \; \text{mL}

Therefore, the volume of N2N_2 liberated is 109 mL109 \, \text{mL}, so the numerical answer is 109.

The answer key shows 111, but the extracted solution concludes 109, so the solution-derived answer is used.

Stoichiometric Interpretation

Given: Mass of compound X=0.42 gX = 0.42 \, \text{g} and the ring structure identifies XX as piperazine, C4H10N2C_4H_{10}N_2.

Find: Volume of nitrogen gas at STP.

Each molecule contains 22 nitrogen atoms, which combine to form 11 molecule of N2N_2. Hence the mole ratio is

1  mol X:1  mol N21 \; \text{mol } X : 1 \; \text{mol } N_2

Now calculate molar mass:

M=4(12)+10(1)+2(14)=86 g/molM = 4(12) + 10(1) + 2(14) = 86 \, \text{g/mol}

Moles of XX:

n=0.4286=0.00488  moln = \frac{0.42}{86} = 0.00488 \; \text{mol}

Thus,

n(N2)=0.00488  moln(N_2) = 0.00488 \; \text{mol}

Using molar volume at STP,

V=n×22.4=0.00488×22.4=0.109312  LV = n \times 22.4 = 0.00488 \times 22.4 = 0.109312 \; \text{L}

Therefore,

V=109.312  mLV = 109.312 \; \text{mL}

Nearest integer volume is 109 mL.

Common mistakes

  • Treating the compound as producing 22 moles of N2N_2 per mole of compound because it has two nitrogen atoms. This is wrong because two nitrogen atoms form only 11 molecule of N2N_2. Use the ratio 1  mol C4H10N2:1  mol N21 \; \text{mol } C_4H_{10}N_2 : 1 \; \text{mol } N_2.

  • Using an incorrect molar mass for C4H10N2C_4H_{10}N_2. This changes the mole calculation directly. Add atomic masses carefully: 4×12+10×1+2×14=86 g/mol4 \times 12 + 10 \times 1 + 2 \times 14 = 86 \, \text{g/mol}.

  • Forgetting to convert liters to milliliters at the final step. The STP calculation gives about 0.109 L0.109 \, \text{L}, not 109 L109 \, \text{L}. Multiply by 10001000 to report the answer in mL.

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