Q. The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of their molecules is:
NTA AbhyasNTA Abhyas 2022
Solution:
Let the mass of oxygen be $x$ gm
$\therefore $ Mass of nitrogen = $4x$ gm
Number of molecules = 6.023 x 1023 x number of moles
$\therefore \text{Ratio} = \frac{\text{6.023} \times 1 0^{2 3} \times \text{no. of moles of oxygen}}{\text{6.023} \times 1 0^{2 3} \times \text{no. of moles of nitrogen}}$
$= \frac{\frac{\text{m}_{\text{O}_{2}}}{\text{M}_{\text{O}_{2}}}}{\frac{\text{m}_{\text{N}_{2}}}{\text{M}_{\text{N}_{2}}}} = \frac{\frac{x}{3 2}}{\frac{4 x}{2 8}}$
$= \frac{7}{3 2} = \text{7 : 32}$
$\therefore $ Mass of nitrogen = $4x$ gm
Number of molecules = 6.023 x 1023 x number of moles
$\therefore \text{Ratio} = \frac{\text{6.023} \times 1 0^{2 3} \times \text{no. of moles of oxygen}}{\text{6.023} \times 1 0^{2 3} \times \text{no. of moles of nitrogen}}$
$= \frac{\frac{\text{m}_{\text{O}_{2}}}{\text{M}_{\text{O}_{2}}}}{\frac{\text{m}_{\text{N}_{2}}}{\text{M}_{\text{N}_{2}}}} = \frac{\frac{x}{3 2}}{\frac{4 x}{2 8}}$
$= \frac{7}{3 2} = \text{7 : 32}$