0.6 g NH2CONH2 (urea) is treated with NaOH and NH3 formed is passed into 300 mL of 0.1 N H2SO4. Volume of unreacted H2SO4 is neutralized by V mL of 1N NaOH.Thus, V is

NH2CONH2+2NaOH⟶2NH3+Na2CO3+H2O2NH3+H2SO4⟶NH42SO4This method is called Kjeldahls' method for estimation of percentage of nitrogenN%=1.4N1 Normality) V1( volume )W( amount ) Percentage of N in urea =2860×100=46.66646.666=1.4×0.1×V1H2SO40.6∴ V1H2SO4=200mLBut, H2SO4 taken = 300 mLUnreacted H2SO4 = 300 - 200 = 100 mL100 mL of 0.1 N H2SO4 = V mL of 1 N NaOH ∴ V=100×0.11=10mL