Verified Solution • 2025–26 Session moderate

QJEEM21Ch6110: JEE Main 2021 Chemistry - Class 11 JEEM21Ch6110 Solution: In Duma's method of estimation of nitrogen, 0.1840 g of an…

In Duma's method of estimation of nitrogen, 0.1840 g of an organic compound gave 30 mL of nitrogen collected at 287 K and 758 mm of Hg pressure. The percentage composition of nitrogen in the compound is __________. (Round off to the Nearest Integer). [Given : Aqueous tension at 287 K = 14 mm of Hg]

Step-by-Step Verified Solution

Aqueous tension at 287 K = 14 mm of Hg.

Hence actual pressure = (758 – 14)
= 744 mm of Hg.

Moles of N2=(758−14)760×30×10−30.0821×287{N_2} = {{\left( {758 - 14} \right)} \over {760}} \times {{30 \times {{10}^{ - 3}}} \over {0.0821 \times 287}}

=1.246×10−3= 1.246 \times {10^{ - 3}} mol

mass of N2=1.246×10−3×28{N_2} = 1.246 \times {10^{ - 3}} \times 28

mass % of N2 =massof′N′totalmass×100

=1.246×28×10−30.184×100= {{1.246 \times 28 \times {{10}^{ - 3}}} \over {0.184}} \times 100

=124.6×280.184%=18.96%= {{124.6 \times 28} \over {0.184}}\% = 18.96\%

≃19%\simeq 19\%

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