×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.