During estimation of nitrogen present in an organic compound by Kjeldahl’s method, the ammonia evolved from 0.5 g of the compound in Kjeldahl’s estimation of nitrogen, neutralized 10 mL of 1 M `H_(2)SO_(4)`. Find out the percentage of nitrogen in the compound.

Explanation

Volume of `H_(2)SO_(4)` used = 10 mL of `1 M H_(2)SO_(4)`
Now `H_(2)SO_(4) + 2NH_(3) rarr (NH_(4))_(2)SO_(4)`
Now 1 mole of `H_(2)SO_(4)` reacts with 2 moles of `NH_(3)`. `:.` 1- mL of `1M H_(2)SO_(4) -= 20 mL` of `1 M NH_(3)`
Now 1000 mL of `1 M NH_(3)` contains `N = 14 g :. 20 mL` of `1 M NH_(3)` will contain `N =(14)/(1000) xx 20`
But this amount of nitrogen is present in 0.5 g of the organic compound.
`:.` Percentage of nitrogen `= (14 xx 20)/(1000) xx (100)/(0.5) = 56.0`.
Alternatively, % age of `N = (1.4 xx "Vol. of acid used" xx "Basicity of acid" xx "Molarity of acid")/("Mass of the substance taken") = (1.4 xx 10 xx 2 xx 1)/(0.5) = 56.0`