JEE Main 2014 — Some Basic Concepts of Chemistry Question with 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
$\therefore \text{Ratio}=\frac{\text{6.023}\times 10^{23}\times \text{no. of moles of oxygen}}{\text{6.023}\times 10^{23}\times \text{no. of moles of nitrogen}}$
              $=\frac{\frac{{\text{m}}_{{\text{O}}_2}}{{\text{M}}_{{\text{O}}_2}}}{\frac{{\text{m}}_{{\text{N}}_2}}{{\text{M}}_{{\text{N}}_2}}}=\frac{\frac{x}{32}}{\frac{4x}{28}}$
              $=\frac{7}{32}=\text{7 : 32}$