# What is the density of O_2 at 298 K and 0.987 atm?

Dec 29, 2016

${\rho}_{{O}_{2}} \cong 1 \cdot g \cdot {L}^{-} 1$

#### Explanation:

PV=nRT; P/(RT)=n/V=("mass"/"molar mass")/V.

And thus $\frac{P}{R T} \times \frac{\text{molar mass"="mass}}{V} = \rho$

So we simply fill in the blanks,

rho=(1*cancel(atm)xx32.00*g*cancel(mol^-1))/(0.0821*L*cancel(atm)*cancel(K^-1)*cancel(mol^-1)xx298*cancel(K)

rho=??*g*L^-1

We get the required units of density.