JEE Main 16 March 2021 Shift 2 Solved Paper

Given, mass of organic compound 0.1840g
Total pressure =758mm of Hg
Aqueous tension =14mm of Hg
Partial pressure of dry N2=758−14
=744mm of Hg
For N2
V=30mL=0.03L
T=287K
p=744mm of Hg=‌

744

760

atm
So, moles of N2(n)=‌

pV

RT

=‌

744

760

×‌

0.03

0.0821×287

n=1.25×10−3 moles of N2
Weight of N2= Moles × Molar mass
=1.25×10−3×28
=35×10−3g
% composition of N2
=‌

‌ Mass of ‌N2

‌ Mass of organic compound ‌

×100
=‌

35×10−3

0.1840

×100
=19.02%