The correct option is C 6.25%
Here,
Mass of organic substance =0.3 g
Volume of Nitrogen collected =50 mL
Temperature (T1)=300 K
Vapour pressure of water =15 mm
Actual pressure of dry nitrogen =715−15=700 mm
At STP, we know that P2=760 mm and T2=273 K then V2=?
⟹P1V1T1=P2V2T2
⟹700×50300=760×V2273
⟹V2=41.9 mL
Hence, % of nitrogen =28×41.9×10022400=5.238%≃6.25%