The relative molecular mass of a volatile liquid can be found by means of a procedure involving the use of either apparatus A or apparatus B shown below - Leaving Cert Chemistry - Question 3 - 2012

To determine the mass of vapour, one must weigh the flask with all fittings before the liquid is vaporized. After the experiment, reweigh the flask to find the difference in mass, which represents the mass of the vapour that has escaped. For the volume, the vapour can be injected into a gas syringe (Apparatus B) or measured using graduated cylinders (Apparatus A). This involves filling the graduated cylinder with a known volume of liquid that once vaporized, the change in volume will reveal the volume of the vapour.

The pressure of the vapour equals atmospheric pressure because when the vapour is in contact with the liquid, it is open to the environment. This allows it to equilibrate, meaning that the vapour pressure adjusts to meet the atmospheric pressure, resulting in equilibrium.

To calculate the number of moles ( ( n )) of vapour, we use the ideal gas equation:

$PV = nRT$

Where:

• ( P = 101 \text{ kPa} = 101000 \text{ Pa} )
• ( V = 0.330 , \text{L} = 330 \text{ cm}^3 = 0.000330 \text{ m}^3 )
• ( R = 8.31 \text{ J/(K mol)} )
• ( T = 100 \degree C = 373 \text{ K} )

Rearranging the equation to find ( n ): $n = \frac{PV}{RT} = \frac{101000 \cdot 0.000330}{8.31 \cdot 373} = 0.01076 \text{ moles}$

Next, to calculate the relative molecular mass (( M_r )): ( M_r = \frac{\text{mass}}{n} = \frac{0.63 , \text{g}}{0.01076 , \text{moles}} = 58.6 , \text{g/mol} ).

This method is unsuitable for non-volatile liquids because these liquids do not vaporize significantly under experimental conditions. As a result, their vapor pressure does not provide reliable data for calculations. Non-volatile liquids would require a different approach, where accurate measurements may involve using techniques such as mass spectrometry.