In the Kjeldahl's methodd for estimation of nitrogen present in a soil sample, ammonia evoled from 0.75 g of sample netralised 10 mL of `1MH_(2)SO_(4)`. The precentage of nitrogen in the soil is-

Options

A. 37.33
B. 45.33
C. 35.33
D. 43.33

Answer

A

Explanation

`H_(2)SO_(4)+2NH_(3)to(NH_(4))_(2)SO_(4)`
`10mL " " 1(M)H_(2)SO_(4)-=10mL " " 2(N)H_(2)SO_(4)`
`-=20mL " " 1(N)H_(2)SO_(4)`
`-=20mL " " 1(N)H_(2)SO_(4)`
`-=20mL " " 1(N)NH_(3)`
`=(20xx17)/(1000)g = 0.34 g NH_(3)`
Amount of N present in 0.34 `NH_(3)=(14xx0.34)/(17)=0.28g`.
So, amount of nitrogen present in 0.75 g of sample = 0.28g.
Hence % of nitrogen in the sample `(0.28xx100)/(0.75) = 37.33`