# How many grams of Kr are in a 2.74 L cylinder at 58.7 °C and 2.99 atm?

Aug 3, 2018

I get a mass of approx. $25 \cdot g$...

#### Explanation:

We use the old Ideal Gas equation...

$n = \frac{P V}{R T} = \frac{2.99 \cdot a t m \times 2.74 \cdot L}{0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \cdot 331.9 \cdot K}$

$= 0.301 \cdot m o l$...

And we know (how) that krypton has a molar mass of $83.8 \cdot g \cdot m o {l}^{-} 1$...

And so we take the product...…..

0.301*molxx83.8*g*mol^-1=??*g