AIEEE 2004 Chemistry Solved Paper

Initially total H2SO4 present =100‌mL of 0.1M=‌

100

1000

×0.1 mole =0.01 mole
2‌NaOH+H2SO4→Na2SO4+2H2O
Let in this reaction H2SO4 required n mole
∴‌

‌ 20 1000 ×0.5

2

=‌

n

1

⇒n=0.005‌ mole ‌
So now remaining H2SO4=0.01−0.005=0.005 mole. Those remaining H2SO4 will react with
NH3‌. ‌
2NH3+H2SO4→(NH4)2SO4
Let no moles of NH3 produce through this reaction is =x
∴‌

x

2

=‌

0.005

1

⇒x=0.01
In NH3 no of N atom is 1 and H atom is 3 . So no of moles of N atom in NH3=0.01×1=0.01 mole
This 0.01 mole or 0.01×14‌gm‌N is produced from 0.3‌gm unknown organic compound.
∴% of N in unknown compound is
‌=‌

0.01×14

0.3

×100
‌=46.6
% of N in urea [(NH4)2‌CO]=‌

14×2

60

×100=46.6%
[Mol weight of urea =60]
% of N in benzamide [C6H5CONH2]=‌

14

121

×100=11.5%
[ Mol weight of benzamide [C6H5CONH2]=121 ]
% of N in acetamide [CH3CONH2]=‌

14

59

×100=23.4%
[ Mol weight of acetamide [CH3CONH2]=59 ]
% of N in thiourea [NH2CONH2]=‌

14×2

76

×100=36.8%
[ Mol weight of thiourea [NH2CSNH2]=76 ]
∴ compound is urea.