Question

A \[\text{135 mL}\] of a gas is collected over water at ${{25}^{\text{0}}}\text{C}$ and $\text{0}\text{.993 bar}$. If the gas weighs $\text{0}\text{.160 gram}$ and the aqueous tension at ${{25}^{\text{0}}}\text{C}$ is $0.0317$, calculate the molecular mass of the gas.

Answer 1

Hint:This question is based on the ideal gas law equation in which the pressure, the volume, the number of moles of the gas, and the temperature of the gas in absolute scale are related to each other by the following equation:
Formula Used: $\text{P V = nRT}$
Where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, and T is the temperature of the gas in absolute scale.

Complete step by step answer:
According to the question it is given that,
Volume of the gas = \[\text{135 mL}\]= $\text{0}\text{.135 L}$
Temperature of the gas =${{25}^{\text{0}}}\text{C}$= $\text{298 K}$
Pressure of the gas = $\text{0}\text{.993 bar}$= $\text{0}\text{.993 atm}$
Weight of the gas = $\text{0}\text{.160 gram}$
Aqueous tension = $0.0317$
The actual pressure of the gas = given pressure - aqueous tension= $\text{0}\text{.993 -0}\text{.0317}$= $0.9613$ atmosphere.
Putting the value of the pressure in the ideal gas equation we get,
$\text{n = }\dfrac{\text{0}\text{.916}\times \text{0}\text{.135}}{0.082\times 298}\text{ }$= $0.005$.
Now, the number of moles of a gas is equal to the mass of the gas by its molecular weight. Therefore, $\text{n = }\dfrac{\text{m}}{\text{M}}$$\Rightarrow \text{M = }\dfrac{0.160}{0.005}$= $\text{30}\text{.12 g/mol}$.

So, the molecular mass of the gas is $\text{30}\text{.12 g/mol}$.
Note:
The Aqueous Tension is defined as the partial pressure of the water vapour present in moist gas. Whenever any gas is collected over water after it is synthesized from a chemical reaction, then the gas becomes moist from the water vapour that has evaporated from the water over which it was collected. The unit of aqueous tension is the same as that of pressure.