# 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?

Jun 3, 2015

Working equation: $P V = n R T$ (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
$3 \cdot 1$ = $\left(\frac{4.37}{M}\right) \cdot 0.082 \cdot 318$ (Temperature in Kelvin scale)
$\implies 3 = \frac{4.37}{M} \cdot 0.082 \cdot 318$
$\implies 3 = \frac{113.95}{M}$
=>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.