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Some Basic Concepts of Chemistry question

2025 · 3 Apr · Shift 2 · Q20
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Some Basic Concepts of Chemistry question

2025 · 3 Apr · Shift 2 · Q20

JEE MainChemistrySome Basic Concepts of ChemistryMCQ+4 / −1
In Dumas' method for estimation of nitrogen 0.4 g of an organic compound gave 60 mL of nitrogen collected at 300 K temperature and 715 mm Hg pressure. The percentage composition of nitrogen in the compound is (Given : Aqueous tension at 300 K=15 mmHg300 \mathrm{~K}=15 \mathrm{~mm} \mathrm{Hg}300 K=15 mmHg )
  1. A
    15.71%15.71 \%15.71%
  2. B
    17.46%17.46 \%17.46%
  3. C
    7.85%7.85 \%7.85%
  4. D
    20.95%20.95 \%20.95%
View written solutionFree

Correct answer: A

  1. Correct the pressure of dry nitrogen

In Dumas' method, the gas collected over water contains water vapour, so pressure of dry N2N_2N2​ is:

PN2=715−15=700 mm HgP_{N_2} = 715 - 15 = 700\ \text{mm Hg}PN2​​=715−15=700 mm Hg

Convert to atm:

PN2=700760 atmP_{N_2} = \frac{700}{760}\ \text{atm}PN2​​=760700​ atm

  1. Use ideal gas equation to find moles of nitrogen gas

Given:

  • Volume, V=60 mL=0.060 LV = 60\ \text{mL} = 0.060\ \text{L}V=60 mL=0.060 L
  • Temperature, T=300 KT = 300\ \text{K}T=300 K
  • Gas constant, R=0.0821 L atm mol−1K−1R = 0.0821\ \text{L atm mol}^{-1}\text{K}^{-1}R=0.0821 L atm mol−1K−1

So,

n=PVRT=(700760)(0.060)0.0821×300n = \frac{PV}{RT} = \frac{\left(\frac{700}{760}\right)(0.060)}{0.0821 \times 300}n=RTPV​=0.0821×300(760700​)(0.060)​

n≈0.0552624.63≈2.24×10−3 mol of N2n \approx \frac{0.05526}{24.63} \approx 2.24 \times 10^{-3}\ \text{mol of } N_2n≈24.630.05526​≈2.24×10−3 mol of N2​

  1. Find mass of nitrogen present

Each mole of N2N_2N2​ has mass 28 g28\ \text{g}28 g.

Mass of nitrogen=n×28\text{Mass of nitrogen} = n \times 28Mass of nitrogen=n×28

=2.24×10−3×28≈0.0628 g= 2.24 \times 10^{-3} \times 28 \approx 0.0628\ \text{g}=2.24×10−3×28≈0.0628 g

  1. Calculate percentage of nitrogen in the compound

Mass of organic compound =0.4 g= 0.4\ \text{g}=0.4 g

%N=0.06280.4×100\%N = \frac{0.0628}{0.4} \times 100%N=0.40.0628​×100

%N=15.7%\%N = 15.7\%%N=15.7%

  1. Match with the options

The closest option is:

15.71%\boxed{15.71\%}15.71%​

So, Option A is correct.

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