(d)
100 ML $O_{2},NH_{3}$ and $CO_{2} = \frac{0.1}{22.4} = \frac{1}{224}$ mol
For $O_{2}$no. of molecules $= \frac{1}{224} \times 6.023 \times 10^{23}$
For $NH_{3}$ no. of molecules $= \frac{1}{224} \times 6.023 \times 10^{23}$
For $CO_{2}$ no. of molecules $= \frac{1}{224} \times 6.023 \times 10^{23}$