Question

16. A 4.37 g sample of a certain diatomic gas occupies a volume of 3.00-L at 1.00 atm and a temperature of 45°C. Identify the gas. Anyone please help?

Answer 1

Working equation: `PV = nRT` (Ideal gas law )
n (no of moles) = m(mass)/M (Molecular wt. )

R in Lt-atm scale 0.08205736(14)L-atm /K-mol ~ 0.082

So, fitting values
`3*1` = `(4.37/M)*0.082*318` (Temperature in Kelvin scale)
`=>3= 4.37/M*0.082*318`
`=>3= 113.95/M`
`=>M ~ 37.985`

Now the only diatomic gas which has closest molecular mass is Fluorine ( `F_2` ) (37.997). The difference in value is probably coming from approximation.