A closed container having a volume of 0.02 m3 contains a mixture of neon and argon at 27^oCand pressure of 1$\times$10⁵ Nm^-2. The mass of mixture is 28 g. Choose the correct options.

$\mathrm{T}={27}^{\circ }\mathrm{C}=300\mathrm{K}$

Let m = mass of neon gas in the mixture, then
mass of argon = 28 - m
Number of gram moles of neon $=\frac{\mathrm{m}}{20}={\mathrm{n}}_1$
Number of gram moles of argon $\frac{28-\mathrm{m}}{40}={\mathrm{n}}_2$
Total pressure of mixture, p = Partial pressure of neon + partial pressure of argon

$\therefore \mathrm{p}={\mathrm{p}}_1+{\mathrm{p}}_2=\frac{{\mathrm{n}}_1\mathrm{RT}}{\mathrm{V}}+\frac{{\mathrm{n}}_2\mathrm{RT}}{\mathrm{V}}$

                          $=\left({\mathrm{n}}_1+{\mathrm{n}}_2\right)\frac{\mathrm{RT}}{\mathrm{V}}$

Substituting the given values, we have

         $1\times {10}^5=\left(\frac{\mathrm{m}}{20}+\frac{28-\mathrm{m}}{40}\right)\left(\frac{8.314\times 300}{0.02}\right)$

$\Rightarrow \mathrm{m}=4.074\mathrm{g}$

So, mass of neon = 4.074 g and mass of argon = 28 - m

                                                                            = 23.926 g