Question #52822

Oct 17, 2016

Let the mass of CO was $w g$

Now

$M \to \text{Molar mass of "CO =28" g/mol}$

$V \to \text{Initial Vol of CO} = 0.52 L$

$P \to \text{Initial Pressure of CO} = 1.5 a t m$

$V \to \text{Initial Temp of CO} = 331 K$

$R \to \text{Universal gas constant} = 0.082 L a t m {K}^{-} 1 m o {l}^{-} 1$

$P V = \frac{w}{M} R T$

$\implies w = \frac{P V M}{R T} = \frac{1.5 \times 0.52 \times 28}{0.082 \times 331} = 0.805 g$

In certain adjusted pressure and temperature when its volume becomes 2.7L ,its density will be
$D = \text{mass"/"volume} = \frac{0.805}{2.7} \frac{g}{L} \approx 0.3 \frac{g}{L}$