If 4g of a gas occupied 11.2L at 0C and 0.25atm calculate molecular mass of the gas.
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We can use ideal gas law here. We have the following data.
`P = 0.25atm`
`V = 11.2L`
`R = 0.08206(atmL)/(molK)`
`T = 273K`
`PV = nRT`
`n = (PV)/(RT)`
`n = (0.25xx11.2)/(273xx0.08206)`
`n = 0.125mol`
If the molecular mass of the gas is M;
`M = (mass)/(mol)`
`M = 4/0.125`
`M = 32g/(mol)`
So the molecular mass of the gas is `32g/(mol)` . Answer is C).
This looks like `O_2` gas.
We assume the gas is an ideal gas.
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