Question

An evacuated glass vessel weighs $50.0{\text{ }}g$ when empty, $148.0{\text{ }}g$ when filled with a liquid of density $0.98{\text{ }}g{\text{ }}m{L^{ - 1}}$ ​, and $50.5{\text{ }}g$ , when filled with an ideal gas at $760{\text{ }}mm{\text{ }}Hg$ at $300{\text{ }}K$. Determine the molar mass of the gas.

Answer 1

In order to answer the given question we will use some known formulas which is the formula of volume to find out the volume of the liquid. After finding the volume of the liquid we will use the ideal gas law equation to find the molecular mass of the gas.

Complete answer: Given,
The weight of an evacuated glass vessel is equal to $50.0{\text{ }}g$
The weight of the liquid = $148.0{\text{ }}g$
Density of liquid = $0.98{\text{ }}g{\text{ }}m{L^{ - 1}}$
It weighs $50.5{\text{ }}g$ when filled with ideal gas at $760{\text{ }}mm{\text{ }}Hg$ and $300{\text{ }}K$ .
The liquid's actual weight is= $148.0{\text{ }} - {\text{ }}50.0{\text{ }} = {\text{ }}98{\text{ }}g.$
Volume of the liquid $= \dfrac{{Mass}}{{Density}}$=$\dfrac{{98}}{{0.98}} = 100$
The vessel volume = $100{\text{ }}ml$ vessel contains "ideal gas" at $760{\text{ }}mm{\text{ }}Hg$ and $300{\text{ }}K$ pressure and temperature, respectively.
Ideal gas weight = $50.5{\text{ }} - {\text{ }}50.0{\text{ }} = {\text{ }}0.5{\text{ }}g$
The ideal gas law is the state equation for a hypothetical ideal gas. Despite its flaws, it is a good approximation to the behaviour of many gases under a variety of situations. $PV{\text{ }} = {\text{ }}nRT$ is the ideal gas equation.
Considering, $PV{\text{ }} = {\text{ }}nRT$
We know that, $n = \dfrac{w}{m}$
So, $PV$= $\dfrac{w}{m}RT$
$\Rightarrow \dfrac{{760}}{{760}} \times \dfrac{{100}}{{1000}} = \dfrac{{0.5}}{M} \times 0.0821 \times 300 \\\ \Rightarrow M = 123.15 \approx 123g \\\$
(We know that the ideal gas pressure is $760mm{\text{ }}Hg$ .) The given pressure is also $760mm{\text{ }}Hg$ . As a result, $\dfrac{{760}}{{760}}{\text{ }} = {\text{ }}P{\text{ }} = {\text{ }}1$ )
As a result, the gas's "molecular weight" $\left( m \right)$ = $123{\text{ }}g\,mo{l^{ - 1}}.$

Note:
The equation of state given here $\left( {PV = nRT} \right)$ only applies to an ideal gas or a real gas that behaves sufficiently like an ideal gas as an approximation. The equation of state can be expressed in a variety of ways. The ideal gas law is most accurate for monatomic gases at high temperatures and low pressures because it ignores both molecular size and intermolecular attractions.