A 5.0 L box contains 28.4 g of mixture of gases ${\text{C}}_{\text{x}}{\text{H}}_{\text{8}}{\text{\hspace{0.17em}and\hspace{0.17em}C}}_{\text{x}}{\text{H}}_{\text{10}}$ . The pressure of the gas mixture at 300 K is 2.46 atm. The analysis of the gas mixture shows that carbon is 84.5% by mass. The value of x is (Take nearest integer) (Assume ideal behaviour)

Let the mixture contains a moles ${\text{C}}_{\text{x}}{\text{H}}_{\text{8}}{\text{\hspace{0.17em}and\hspace{0.17em}b\hspace{0.17em}moles\hspace{0.17em}C}}_{\text{x}}{\text{H}}_{\text{10}}$ .
 Now,  $\text{a×}\left(\text{12x+8}\right)\text{+b×}\left(\text{12x+10}\right)\text{=28.4\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\text{1}\right)$
  $\text{a+b=}\frac{PV}{RT}=\frac{2.46\times 5}{0.082\times 300}=0.5\left(2\right)$
 And  $28.4\times \frac{84.5}{100}=a\times 12x+b\times 12x\left(3\right)$
On solving,  $x\approx 4$