# A 283.3*g mass of an elemental gas X exerts a pressure of 3.2*atm at a temperature of 300*K, while it is enclosed in a 30*L volume. What is the identity of the element?

May 10, 2017

$\text{Element X} \equiv C l$

#### Explanation:

We use the ideal gas equation..........

$n = \frac{P V}{R T} = \frac{3.2 \cdot a t m \times 30 \cdot L}{0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 300 \cdot K}$

$= 3.90 \cdot m o l$

But since we KNOW the mass of this gas;

$\text{Molar mass"="Mass of gas"/"Molar quantity of gas}$

$\frac{283.3 \cdot g}{3.90 \cdot m o l} = 72.6 \cdot g \cdot m o {l}^{-} 1$

Since we know that this is a binuclear element, i.e. ${X}_{2}$, the atomic mass of the element is approx. $36 \cdot g \cdot m o {l}^{-} 1$. Chlorine is the likely candidate, $\text{atomic mass} = 35.45 \cdot g \cdot m o {l}^{-} 1$. And note that chlorine, like most of the other elemental gases is BINUCLEAR, i.e. ${X}_{2}$.