29.5 MG app in organic compound containing nitrogen was digested according to the house method and the award Ammonia was observed in 20 ml the writer of 0.1 molarity of HCL solution the axis of the acid required 15 ml of 0.1 molarity of a new OS for complete neutralization the percentage of Nitrogen in the compound is

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15 ml and 0.1 M is required for the complete neutralization tehrfore moles that dissolved te ammonia fully would be => 
20*0.1 – 15*0.1 => 2- 1.5 => 0.5mmoles.
Therfore 0.5 Moles of NH3 would have 0.5 mmoles of N and the weight would be => 0.5*14/1000 => 0.007 gms.
Therefore percentage of N in 29.5 mg of organic compound would be => 0.007*100/0.0295 = 23.72%