Some Basic Principles NEET 2011: Dumas Method Solution

What is the answer to the Some Basic Principles NEET 2011 Dumas-method question?

Option A, 16.45%, is the keyed answer to the Some Basic Principles NEET 2011 nitrogen-estimation question. Use this sequence: dry pressure, STP volume, nitrogen mass, then percentage by mass.

An organic compound weighing 0.35 g gives 55 mL of moist nitrogen at 300 K and 715 mm Hg. The water-vapour pressure at this temperature is 15 mm Hg. Find the percentage of nitrogen by mass.

The supplied choices are:

The chapter is Some Basic Principles of Organic Chemistry. The topic is quantitative analysis by Dumas method.

How do you find the pressure of dry nitrogen?

Subtract the water-vapour pressure from the measured pressure. Aqueous tension means the pressure contributed by water vapour at the collection temperature. The collected gas is moist, so the stated pressure includes both nitrogen and water vapour.

Dalton’s law gives:

P_{total} = P_{N_{2}} + P_{water vapour}

Rearrange before substituting:

P_{dry} = P_{total} - P_{water vapour}

P_{dry} = (715 - 15) mm Hg = 700 mm Hg

Use this dry pressure in the gas-volume correction. Aqueous tension is a pressure, so subtracting 15 from the volume or temperature is a unit error. The subtraction removes only the water-vapour contribution to the measured pressure.

How do you convert the measured 55 mL to STP?

Correct for both pressure and temperature to obtain approximately 46.1 mL. The STP convention used in this solution is 273 K and 760 mm Hg. The original temperature is already in kelvin, so no Celsius-to-kelvin conversion is needed.

For the same amount of nitrogen:

\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{0} V_{STP}}{T_{0}}

Rearranging gives:

V_{STP} = \frac{P_{1} V_{1} T_{0}}{P_{0} T_{1}}

The starting values are:

P_{1} = 700 mm Hg, V_{1} = 55 mL, T_{1} = 300 K

Substitute them:

V_{STP} = \frac{700 \times 55 \times 273}{760 \times 300} mL \approx 46.1 mL

Both correction ratios are below one:

\frac{700}{760} < 1, \frac{273}{300} < 1

Raising pressure compresses the gas, while lowering temperature also reduces its volume. Both corrections must therefore make the STP volume smaller than the measured volume.

Use this direction check:

V_{STP} < 55 mL

A larger result signals an error, such as an inverted ratio. A smaller result passes this check but still needs the numerical calculation.

How does the STP volume give 16.45% nitrogen?

Convert the corrected volume to nitrogen mass, then divide by the original sample mass. At the STP convention used here, 22400 mL of molecular nitrogen corresponds to 28 g.

Set up the proportion with matching units, using the rounded volume:

\frac{m_{N_{2}}}{28 g} \approx \frac{46.1 mL}{22400 mL}

m_{N_{2}} \approx \frac{28 \times 46.1}{22400} g \approx 0.0576 g

The molar mass used is:

M (N_{2}) = 28 {g mol}^{- 1}

The gas is molecular nitrogen, and all its mass is nitrogen. Do not halve this mass just because each molecule contains two atoms.

Now calculate the percentage:

% N = \frac{mass of nitrogen}{mass of original compound} \times 100

% N \approx \frac{0.0576}{0.35} \times 100 \approx 16.45 %

The supplied official result is 16.45%, option A; the intermediate values shown are rounded. Reuse this sequence: subtract aqueous tension, correct to STP, convert to mass, divide by sample mass.

Can skipping the temperature correction lead you to option B?

Only by forcing a nearest-option choice: the calculation does not produce B’s value. Keeping the correct dry pressure of 700 mm Hg while omitting the temperature factor gives approximately 18.09%, not 17.45%. That mismatch is not a rounding error.

Here is the deliberately faulty pressure-only correction:

V_{wrong} = \frac{700 \times 55}{760} mL \approx 50.66 mL

Continuing the same conversion route gives:

% N_{wrong} \approx \frac{28 \times 50.66}{22400 \times 0.35} \times 100 \approx 18.09 %

Among the supplied options, B is nearest to this wrong result. Selecting it would involve two separate mistakes:

Treating a pressure correction as a complete STP conversion.
Using the options to excuse a substantial numerical mismatch.

This is a reproducible wrong route, not the proven origin of option B or a measured student-response pattern. The original gas temperature is not the STP temperature, so pressure correction alone is incomplete.

Restore the omitted factor below and recalculate rather than forcing agreement with an option: $\frac{273}{300}$

Which three practice questions test the same Dumas-method skills?

Test pressure interpretation, molecular mass and the sample-mass denominator. These are original related practice questions, not additional verified PYQs. Each changes one part of the reasoning while keeping the same nitrogen-estimation method.

Should you subtract aqueous tension if dry nitrogen pressure is already given?

No: the water-vapour contribution has already been excluded. Use the stated dry pressure directly in the gas equation. Subtracting aqueous tension again would make the pressure, corrected volume and calculated nitrogen mass too small.

Why do you use 28 rather than 14 in the mass calculation?

The collected gas is molecular nitrogen. One mole of that gas weighs 28 g, whereas 14 g corresponds to one mole of nitrogen atoms.

M (N_{2}) = 2 \times 14 = 28 {g mol}^{- 1}

Using 14 with the molar gas volume would halve the calculated nitrogen mass. With the sample mass unchanged, the nitrogen percentage would also halve.

What happens if twice the sample mass gives the same nitrogen volume?

The nitrogen percentage halves. The same gas volume under identical collection conditions means the same nitrogen mass. Only the sample-mass denominator doubles.

% N_{new} = \frac{m_{N}}{2 m_{sample}} \times 100 = \frac{1}{2} % N_{original}

Now retry the original question without looking at the solution. Use 120 seconds, the question bank’s expected solving time, not an official examination time limit. Write the dry pressure, STP volume, nitrogen mass and percentage as four separate steps so you can locate any error.

Next step: photograph a doubt on NEET JEEnius AI and photograph any question you are stuck on and get a step-by-step solution across Physics, Chemistry and Biology (20 free a month).

For a worked example of the same idea, see Electrostatics NEET 2009: Concentric Shell Potentials.

Frequently asked questions

What is the answer to the NEET 2011 Dumas-method question?

The keyed answer is option A, 16.45% nitrogen by mass. Subtract water-vapour pressure, convert the nitrogen volume to STP, calculate its mass and divide by the original 0.35 g sample mass to obtain the percentage.

How do you subtract aqueous tension in the Dumas method?

Subtract aqueous tension, or water-vapour pressure, from the total pressure of the moist gas. In this question, dry nitrogen pressure is 715 − 15 = 700 mm Hg. If dry nitrogen pressure is already given, do not subtract aqueous tension again.

How do you convert 55 mL of nitrogen at 300 K to STP?

Using the dry pressure of 700 mm Hg and STP conditions of 273 K and 760 mm Hg, calculate V_STP = (700 × 55 × 273)/(760 × 300) ≈ 46.1 mL. Both pressure and temperature corrections reduce the volume, so the result must be below 55 mL.

Why is 28 used instead of 14 in Dumas-method calculations?

The collected gas is molecular nitrogen, N₂, whose molar mass is 28 g/mol. At the STP convention used here, 22400 mL of N₂ corresponds to 28 g. Using 14 g would incorrectly halve the calculated nitrogen mass and percentage.