# What is the likely identity of a gas at 273.15 K, whose mass is 7.5*g, and which exerts a pressure of 1*"bar"?

Jun 19, 2017

Well, let's try to address the molar mass of the gas........we get a range of possibilities.........

#### Explanation:

$\text{STP}$ specifies a temperature of 273.15 K, and an absolute pressure of exactly 100* kPa, "1 bar".

$n = \frac{P V}{R T} = \left(1 \cdot \cancel{\text{bar"xx5.6*cancelL)/(8.314xx10^(-2)cancelL*cancel"bar}} \cdot \cancel{{K}^{-} 1} \cdot m o {l}^{-} 1 \times 273.15 \cdot \cancel{K}\right)$

$= 0.247 \cdot {\left(m o {l}^{-} 1\right)}^{-} 1 = 0.247 \cdot m o l$

And thus $\frac{7.5 \cdot g}{0.247 \cdot m o l} = 30.3 \cdot g \cdot m o {l}^{-} 1$.

The gas is likely the homonuclear diatomic ${N}_{2}$ or ${O}_{2}$.

Further data are needed to confirm the identity of the gas.