Q.In Dumas' method for estimation of nitrogen, 0.3 g of an organic compound gave 50 mL of nitrogen collected at 300 K temperature and 715 mm pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300 K = 15 mm)

Concept & Intuition

We have two separate pieces of information: the combustion percentages (C, H, N, O) and a Kjeldahl titration that tells us the exact nitrogen content. The trick is that the percentages must add to 100%, so once we find N% from the titration, O% = 100 – (C% + H% + N%). Then we convert each percentage to moles of atoms, find the simplest whole‑number ratio, and match it to the given options.

Step‑by‑step reasoning

1. Find the moles of NH₃ (and thus N) from the Kjeldahl titration
   • The ammonia evolved is absorbed in 50 mL of 0.05 M H₂SO₄. Moles of H₂SO₄ taken = $0.050 \, \text{L} \times 0.05 \, \text{mol/L} = 0.0025 \, \text{mol}$.
   • Each mole of H₂SO₄ neutralizes 2 moles of NH₃:

• The excess H₂SO₄ is titrated with 25 mL of 0.1 M NaOH. Moles of NaOH used = $0.025 \, \text{L} \times 0.1 \, \text{mol/L} = 0.0025 \, \text{mol}$. Since 2 NaOH neutralize 1 H₂SO₄, moles of excess H₂SO₄ = $0.0025 / 2 = 0.00125 \, \text{mol}$.
• Therefore, moles of H₂SO₄ that reacted with NH₃ = $0.0025 - 0.00125 = 0.00125 \, \text{mol}$.
• Moles of NH₃ produced = $2 \times 0.00125 = 0.0025 \, \text{mol}$.
• Each mole of NH₃ contains 1 mole of N, so moles of N in the 0.30 g sample = 0.0025 mol.

2. Calculate the percentage of nitrogen in the compound

So $x = 11.67\%$.

3. Find the oxygen percentage by difference Total = 100% = 69.4% C + 5.8% H + 11.67% N + y% O

4. Convert percentages to moles of atoms per 100 g of compound

   • Moles of C = $\frac{69.4}{12} = 5.783$
   • Moles of H = $\frac{5.8}{1} = 5.8$
   • Moles of N = $\frac{11.67}{14} = 0.8336$
   • Moles of O = $\frac{13.13}{16} = 0.8206$

5. Find the simplest whole‑number ratio

   Divide each by the smallest number (≈0.8206):

   • C: $5.783 / 0.8206 \approx 7.05$ → 7
   • H: $5.8 / 0.8206 \approx 7.07$ → 7
   • N: $0.8336 / 0.8206 \approx 1.016$ → 1
   • O: $0.8206 / 0.8206 = 1$ → 1

   The ratio is C₇H₇NO.

6. Match with the options

   Option (C) is C₇H₇NO.

1. Find the mass of methane actually burnt.

   You start with 160 g of methane, but 5% remains unburnt. That means only 95% of it participates in the reaction.

   Mass burnt = $160 \times \frac{95}{100} = 152$ g.

2. Convert that mass to moles of methane.

   Molar mass of CH₄ = $12 + 4 \times 1 = 16$ g/mol.

   Moles of CH₄ burnt = $\frac{152}{16} = 9.5$ mol.

3. Write the balanced combustion reaction.

   Methane burns in oxygen:

From the stoichiometry, 1 mole of CH₄ requires 2 moles of O₂.

So, moles of O₂ needed = $9.5 \times 2 = 19$ mol.

4. Find the volume of O₂ at STP. At STP (273 K, 1 bar), the molar volume is given as 22.71 L/mol. Volume of O₂ = $19 \times 22.71 = 431.49$ L.

5. Relate O₂ volume to air volume. Air contains 20% O₂ by volume. That means every 100 L of air contains 20 L of O₂. So, volume of air = $\frac{\text{volume of O}_2}{0.20} = \frac{431.49}{0.20} = 2157.45$ L.

The key concept here is that the "amount of solute" in a solution is measured in moles, not in grams or in any property that depends on the molar mass. Molarity (M) tells us moles per liter, so the total moles of solute is simply:

The molar masses given (106 u, 142 u, etc.) are irrelevant for comparing amounts — they would matter if we were comparing mass of solute, but the question asks for "amount," which in chemistry means moles. Let’s compute each option.

1. Option (A): 1.0 L of 0.25 M Na₂CO₃

2. Option (B): 0.25 L of 0.2 M Na₂SO₄

3. Option (C): 0.5 L of 1.0 M KMnO₄

4. Option (D): 0.75 L of 0.5 M (NH₂)₂CO

Comparing: 0.5 mol (C) > 0.375 mol (D) > 0.25 mol (A) > 0.05 mol (B). So option (C) has the highest amount of solute.

The key idea is that "burning" here means thermal decomposition: CaCO₃ breaks down into CaO and CO₂. The reaction is:

Every mole of CaCO₃ that decomposes yields exactly one mole of CO₂ gas. So the problem reduces to: how many moles of pure CaCO₃ are actually present in the sample, and then how many litres of CO₂ does that produce at STP?

1. Find the mass of pure CaCO₃. The sample is 90% pure by mass. That means out of 10 g of the impure sample, only 90% is actually CaCO₃.

2. Convert that mass to moles. The molar mass of CaCO₃ is given as 100 u, so 1 mol = 100 g.

3. Use the 1:1 stoichiometry.

   From the balanced equation, 1 mol CaCO₃ → 1 mol CO₂. Therefore, moles of CO₂ produced = 0.09 mol.

4. Convert moles of CO₂ to volume at STP.

   The molar volume at STP is given as 22.7 L/mol (not the usual 22.4 L, so use the given value).

Rounding to three decimal places gives 2.044 L.

Concept

For an ideal gas the density fixes the molar mass through $PM=dRT$. For a two-gas mixture the "$M$" that appears is the mole-fraction–weighted average molar mass, so once we compute $M_{avg}$ we can back out the composition.

Step 1 — Average molar mass from the density

Step 2 — Relate $M_{avg}$ to the mole fraction of $N_2$

Let $x$ be the mole fraction of $N_2$ (so $1-x$ is that of $O_2$), with $M_{N_2}=28$ and $M_{O_2}=32$:

Step 3 — Solve for $x$

The mixture ($30.176$) is just above the plain average of $30$, so it is slightly $O_2$-rich, i.e. $x_{N_2}<0.5$ — consistent with $0.456$.

Working.

• Molar mass of $\text{C}_3\text{H}_8\text{O}$: $3(12) + 8(1) + 16 = 36 + 8 + 16 = 60\ \text{g mol}^{-1}$.
• Mass of carbon in one mole: $3\times12 = 36\ \text{g}$.
• Percentage of carbon: $\dfrac{36}{60}\times100 = 60\%$.

Among the four choices, only $60\%$ corresponds to a species of composition $\text{C}_3\text{H}_8\text{O}$ (for example propan-1-ol), consistent with the official key.

We start with the concept: The first ionization energy (IE₁) of Mg is 750 kJ/mol — that’s the energy needed to remove one electron from each Mg atom to form Mg⁺. The second ionization energy (IE₂) is 1500 kJ/mol — the energy to remove a second electron from each Mg⁺ to form Mg²⁺. If we have a fixed amount of Mg and a fixed energy supply, we first use energy to make as many Mg⁺ as possible, and then, if energy remains, we convert some Mg⁺ to Mg²⁺. The trick is to work in moles, not grams.

1. Convert mass to moles.

   Molar mass of Mg = 24 g/mol.

   So 1 g of Mg = $\frac{1}{24}$ mol ≈ 0.04167 mol.

2. Energy needed to fully ionize all Mg to Mg⁺.

   IE₁ = 750 kJ/mol.

   Energy for all Mg → Mg⁺ = $0.04167 \times 750 = 31.25$ kJ.

3. Energy left after forming Mg⁺.

   Total energy absorbed = 50 kJ.

   Remaining energy = $50 - 31.25 = 18.75$ kJ.

4. Use remaining energy to convert Mg⁺ to Mg²⁺.

   IE₂ = 1500 kJ/mol.

   Moles of Mg⁺ that can be further ionized = $\frac{18.75}{1500} = 0.0125$ mol.

5. Determine final amounts.

   • Mg²⁺ formed = 0.0125 mol.
   • Mg⁺ remaining = original moles of Mg⁺ (0.04167) minus those converted (0.0125) = 0.02917 mol.

6. Convert to percentages.

   Total moles of Mg atoms = 0.04167.

   % Mg⁺ = $\frac{0.02917}{0.04167} \times 100 \approx 70\%$.

   % Mg²⁺ = $\frac{0.0125}{0.04167} \times 100 \approx 30\%$.

Concept — equivalence of electrons transferred. In a redox titration, (moles of reductant) × (electrons lost each) = (moles of oxidant) × (electrons gained each).

Step 1 — electrons lost by ferrous oxalate. FeC$_2$O$_4$ contains two oxidisable parts:

• $\mathrm{Fe^{2+} \to Fe^{3+} + e^-}$ → 1 electron
• $\mathrm{C_2O_4^{2-} \to 2CO_2 + 2e^-}$ → 2 electrons

Total = 3 electrons per mole of FeC$_2$O$_4$.

Step 2 — electrons gained by permanganate in acidic medium.

→ 5 electrons per mole of KMnO$_4$.

Step 3 — balance. If $x$ = moles of FeC$_2$O$_4$ oxidised by 1 mol KMnO$_4$:

(The balanced equation is $3\,\mathrm{MnO_4^-} + 5\,\mathrm{FeC_2O_4} + 24\,\mathrm{H^+} \to 3\,\mathrm{Mn^{2+}} + 5\,\mathrm{Fe^{3+}} + 10\,\mathrm{CO_2} + 12\,\mathrm{H_2O}$ — five oxalates per three permanganates, i.e. $5/3$ per one.)

ANSWER: D

The key here is that the air bag undergoes adiabatic expansion — no heat exchange with the surroundings. For an ideal gas under adiabatic conditions, the relation between pressure and volume is given by $PV^\gamma = \text{constant}$, where $\gamma = 1.4$ for air. This is a power-law relationship, so small percentage changes in volume translate into proportional percentage changes in pressure, scaled by $\gamma$.

When you see a problem asking for a percentage change in one variable given a small percentage change in another, the most efficient approach is to use logarithmic differentiation. Instead of solving for exact values, you differentiate the relation to get a direct link between fractional changes. This works beautifully for small changes, which is exactly what "5% increase" implies.

Let’s walk through it step by step.

1. Start with the adiabatic relation For an adiabatic process:

Taking natural logarithms on both sides:

2. Differentiate to relate small changes Differentiate the equation (treating $P$ and $V$ as variables, $k$ constant):

This gives:

The negative sign tells us that pressure and volume change in opposite directions — as volume increases, pressure decreases.

3. Plug in the given percentage change The volume increases by 5%, so:

With $\gamma = 1.4$:

4. Interpret the result A fractional change of $-0.07$ means a 7% decrease in pressure. The question asks for the "percentage change in pressure" — by convention, this is the magnitude of the change (the absolute value), unless a sign is explicitly requested. So the answer is 7%.

The problem is a classic stoichiometry-with-efficiency calculation. Efficiency means that only 80% of the magnetite you put in actually reacts to give iron — the rest is wasted. So you need more magnetite than the ideal stoichiometric amount.

Step 1: Write the balanced chemical equation.

Magnetite is $\text{Fe}_3\text{O}_4$. It is reduced by CO to iron metal and carbon dioxide. The unbalanced reaction is:

Balancing: each Fe$_3$O$_4$ gives 3 Fe atoms, so we need 3 Fe on the right. The oxygen: Fe$_3$O$_4$ has 4 O atoms; each CO takes one O to become CO$_2$, so we need 4 CO molecules. That gives 4 CO$_2$ molecules. The balanced equation is:

Step 2: Find the stoichiometric mass of magnetite for 4 kg of iron.

Molar mass of Fe = 56 g/mol. Molar mass of Fe$_3$O$_4$ = $3 \times 56 + 4 \times 16 = 168 + 64 = 232$ g/mol.

From the equation: 1 mol Fe$_3$O$_4$ produces 3 mol Fe.

Mass ratio: 232 g Fe$_3$O$_4$ gives $3 \times 56 = 168$ g Fe.

So for 4 kg = 4000 g of Fe, the ideal mass of magnetite needed is:

Simplify: divide numerator and denominator by 8:

So:

Step 3: Account for 80% efficiency.

Efficiency = 80% = 0.80 means that only 80% of the magnetite you use actually converts to iron. So the actual mass you need is larger:

Rounding to one decimal place gives 6.9 kg.