40 gms each of the methane and oxygen are mixed in an empty container maintained at 40°C. The fraction of the total pressure exerted by oxygen is

ANS:B - 1/3

To solve this problem, we can use Dalton's law of partial pressures, which states that in a mixture of non-reacting gases, the total pressure exerted by the mixture is equal to the sum of the partial pressures of individual gases. First, let's find the moles of each gas:

1. Moles of methane (CH₄): Molar mass of CH4=12 g/mol+4×1 g/mol=16 g/molMolar mass of CH4​=12g/mol+4×1g/mol=16g/mol Number of moles of CH4=40 g16 g/mol=2.5 molNumber of moles of CH4​=16g/mol40g​=2.5mol
2. Moles of oxygen (O₂): Molar mass of O2=2×16 g/mol=32 g/molMolar mass of O2​=2×16g/mol=32g/mol Number of moles of O2=40 g32 g/mol=1.25 molNumber of moles of O2​=32g/mol40g​=1.25mol

Now, let's find the partial pressures of each gas using the ideal gas law: 𝑃𝑉=𝑛𝑅𝑇PV=nRT

1. For methane: 𝑃CH4=𝑛CH4𝑅𝑇𝑉PCH4​​=VnCH4​​RT​
2. For oxygen: 𝑃O2=𝑛O2𝑅𝑇𝑉PO2​​=VnO2​​RT​

Since both gases are in the same container and at the same temperature, we can compare their pressures directly. The ratio of the partial pressure of oxygen to the total pressure is: 𝑃O2𝑃total=𝑛O2𝑛CH4+𝑛O2Ptotal​PO2​​​=nCH4​​+nO2​​nO2​​​ Substituting the values: 𝑃O2𝑃total=1.252.5+1.25=1.253.75=13Ptotal​PO2​​​=2.5+1.251.25​=3.751.25​=31​ So, the fraction of the total pressure exerted by oxygen is 1331​. Therefore, the correct answer is 1/3.