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

Answer:

The gas was Hexane

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:

$d_{g} = \frac{0.3665 g}{0.1040L} = 3.524 g/L$

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

$d_{g} = \frac{P*M}{R*T}$

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

$M=\frac{d_{g}*R*T}{P}=\frac{3.524g/L*0,082atmL/molK*291K}{0.976atm}$

$M=86.16 g/mol$

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

The gas with the closer molar mass is Hexane