Question

A sample of an unknown volatile liquid was injected into a Dumas flask (mflask = 27.0928 g, Vflask = 0.1040 L) and heated until no visible traces of the liquid could be found. The flask and its contents were then rapidly cooled and reweighed (mflask+vapor = 27.4593 g). The atmospheric pressure and temperature during the experiment were 0.976 atm and 18.0 °C, respectively. The unknown volatile liquid was ________.

Answer 1

The gas was Hexane

Step-by-step explanation:

taking the diference between the mass of the flask and the final mass qe can calculate the mass of liquid injected (assuming none escaped the flask):

`m_(l)  = 27.4593g - 27.0928g = 0.3665g`

with the volume of the flask we can get the density of the gas at the indicated pressure and temperature:

`d_(g)  = (0.3665 g)/(0.1040L) = 3.524 g/L`

From the ideal gases law we have that the density can be calculated as:

`d_(g)  = (P*M)/(R*T)`

Where R is the ideal gases constant = , and M the molecular weight of the fluid. Solving for M:

`M=(d_(g)*R*T)/(P)=(3.524g/L*0,082atmL/molK*291K)/(0.976atm)`

`M=86.16 g/mol`

Note that the temperature is computed in Kelvin T= 18+273=291K

The gas with the closer molar mass is Hexane