The correct option is D 72%
Let the volume of NH3 be x ml, then the volume of H2 would be (50−x) ml.
On sparking,
NH3(g)→(1/2) N2(g)+(3/2) H2(g)
Since 40 ml of O2 is added and sparked, it must have reacted with H2 to form liquid water. Since 6 ml contraction in volume is there with alkaline pyrogallol as it is known to absorb O2 , so 34 ml is the volume of O2 used up.
∴ Total volume of H2 would be 68 ml.(∵2H2+O2→2H2O)
∴(50−x)+32x=68,50+x2=68,x=36 ml
∴ % volume of NH3 in the original mixture =3650×100=72%