Question

0.5 grams of an organic compound liberate 112ml of nitrogen at stp the percentage of nitrogen in compound is

Answer 1

Answer:

14 percent  

Explanation:

we know that at stp 1 mole gives 22400 ml

here we know the liberated amount is 112 ml so the no of moles of nitrogen present is 112/22400=0.005 moles

1 mole of nitrogen weighs 14 gms then 0.005 moles weighs 0.07 so the percentage of nitrogen from the total mass is given as

0.07/0.5 ×100 =14 percent