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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 $=4 x\, gm$
Number of molecules $=6.023 \times 10^{23} \times$ number of moles
$\therefore \text { Ratio }=\frac{6.023 \times 10^{23} \times \text { no. of moles of oxygen }}{6.023 \times 10^{23} \times \text { no. of moles of nitrogen }} $
$=\frac{\frac{ m _{ O _2}}{ M _{ O _2}}}{\frac{ m _{ N _2}}{ M _{ N _2}}}=\frac{\frac{x}{32}}{\frac{3 x}{28}} $
$=\frac{7}{32}=7: 32$