Vikas TU
Last Activity: 7 Years ago
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%