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 ________.

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ramirocarracedo ramirocarracedo · Answer 1 · 2019-10-24T04:06:36+02:00

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):

[tex]m_{l} = 27.4593g - 27.0928g = 0.3665g[/tex]

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

[tex]d_{g} = \frac{0.3665 g}{0.1040L} = 3.524 g/L[/tex]

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

[tex]d_{g} = \frac{P*M}{R*T}[/tex]

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

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

[tex]M=86.16 g/mol[/tex]

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

The gas with the closer molar mass is Hexane