Determination of Molecular Mass Victor-Meyer's Method
In this method a known mass of a volatile liquid or solid is converted into its vapour by heating in a Victor-Meyer's tube. The vapour displaces its own volume of air. The volume of air displaced by the vapour is measured at the experimental temperature and pressure. The volume of the vapour at s.t.p is then calculated. From this the mass of 2.24 x 10-2m3 of the vapour at S.T.P. is calculated. This value represents the molecular mass of the substance.
The apparatus consists of an inner Victor-Meyer tube, the lower end of which is in the form of a bulb. The upper end of the tube has a side tube which leads to a trough of water. The Victor-Meyer tube is surrounded by an outer jacket. In the outer jacket is placed a liquid which boils at a temperature at least 30 K higher than the boiling point of the volatile substance under study. A small quantity of glass wool or asbestos fiber covers the bottom of the Victor-Meyer tube to prevent breakage when the bottle containing the substance is dropped in.
The liquid in the outer jacket is allowed to boil and when no more air escapes from the side tube, a graduated tube filled with water is inverted over the side tube dipping in a trough full of water. A small quantity of the substance is exactly weighed in a small stoppered bottle and quickly dropped in the heated Victor-Meyer tube and corked immediately.
The bottle falls on the asbestos pad and its content suddenly changes into vapour, blow out the stopper and displace an equal volume of air which collects in the graduated tube. The volume of air in the graduated tube is measured by taking it out by closing its mouth with the thumb and dipping it in a jar full of water. When the water levels outside and inside the tube are the same, the volume of air displaced is noted. The atmospheric pressure and laboratory temperature are noted.
Mass of the volatile substance = wg
Volume of air displaced = Volume of vapour = V1 m3
Laboratory temperature = T1 K
Let the atomospheric pressure be P
Pressure of dry vapour = Atomospheric pressure - aqueous tension at. T1 K Let the aqueous tension be p Nm-2 at that temperature.
Pressure of dry vapour = P1 = [P-p]
Standard temperature = T0 = 273 K
Standard pressure = = P0 = = 1.013 x 105 Nm-2
Let the volume of the vapour at standard temperature and pressure be V0 m3
From the gas equation, it follows
P0V0 / T0 = P1 V1 / T1
V0 = (P1 V1 / T1) x (T0/ P0)
The mass of V0 m3 of vapour at s.t.p is w g.
The mass of 2.24 x 10-2 m3 of the vapour at s.t.p. is
= 2.24x10-2xW / V0
The value thus calculatd gives the molecular mass
Molecular mass = 2 x vapour density
Vapour density = Molecular mass / 2
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