What volume of air contains 10.0g of oxygen gas at 273 k and 1.00 atm?

The volume of air will be 7.00 L

Explanation:

To calculate the number of moles, we use the equation:

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in above equation, we get:

To calculate the volume of gas, we use the equation given by ideal gas that follows:

where,

P = pressure of the gas  = 1.00 atm

V = Volume of the gas  = ?

T = Temperature of the gas = 273 K

R = Gas constant =

n = number of moles of gas = 003125 mol

Putting values in above equation, we get:

Hence, the volume of air will be 7.00 L

the answer is option a.