In Duma's method for estimation of nitrogen

Explanation :

Given data:

                   m = 0.25 g
                   V1 = 40 ml
                   T1 = 300 K
                   P1 = 700 mm(725 mm - 25 mm)

                   P0 = 760 mm, T0 = 273 K, V0 = ?

22400 ml of N2 at STP weighs = 28 gm

Mass of nitrogen at 33.53 ml is-

                    = 28 gm  33.53 ml22400 ml

                    = 0.0419 gm

Hence, the percentage of nitrogen is-

                    = 0.0419 gm  1000.25 gm

                    = 16.76 %

Final Answer:
Hence, Option (A) is correct i.e. 16.76