# If 4g of a gas occupied 11.2L at 0C and 0.25atm calculate molecular mass of the gas. A)8 B)16 C)32 D)48 E)64

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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.**

*Note*

*We assume the gas is an ideal gas.*

**Sources:**