• # question_answer Equal amounts of two gases of molecular weight 4 and 40 are mixed. The pressure of the mixture is 1.1 atm. The partial pressure of the light gas in this mixture is [CBSE PMT 1991] A)                 0.55 atm              B)                 0.11 atm C)                 1 atm    D)                 0.12 atm

No. of moles of lighter gas $=\frac{m}{4}$                    No. of moles of heavier gas $=\frac{m}{40}$                    Total no. of moles $=\frac{m}{4}+\frac{m}{40}=\frac{11m}{40}$                    Mole fraction of lighter gas $=\frac{\frac{m}{4}}{\frac{11m}{40}}=\frac{10}{11}$                    Partial pressure due to lighter gas $={{P}_{o}}\times \frac{10}{11}$                                 $=1.1\times \frac{10}{11}=1atm.$