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The Experimental Determination of the Molar Mass of a Gas 1) The air in a flask

ID: 1040523 • Letter: T

Question

The Experimental Determination of the Molar Mass of a Gas

1) The air in a flask is to be replaced with He gas in a way similar to how air is replaced in the experiment. Should the flask be (A) hold upside down, (B) hold upright, (C) either way or (D) not enough information to answer? Provide an explanation below. Molar mass of air is ~ 29 g/mol.

2) A sample of a gas has a mass of 0.293 g and a pressure of 600.0 mm Hg at 14.0 °C in a 100.0 mLcontainer. What is the molar mass of the gas?

3) List all possible sources of error that could occur during this experiment when determining molar mass of a gas experiments?

4) In the experiment, if the collected gas CH4 were contaminated with water vapor, would the experimental molar mass of CH4 be (A) higher, (B) lower, (C) no change or (D) not enough information to answer? Assuming the mistake mentioned was the only mistake in the experiment. Provide an explanation in the space below.

5) why is it necessary to determine the mass of the empty flask when performing a molar mass of gas experiment?

Explanation / Answer

1) He is less density than air. Best example filling of baloon with Helium.

2) PV=nRT

P = 0.7894 atm, V = 0.1 L, T = 287, R = 0.0821 L atm mol-1 K-1, n= wt/ m.wt =0.293/X

substituting above value in eqn 0.7894x0.1=(0.293/M.wt)x0.0821x287

3) changing the volume of the gas effect the preasure that may error.

4) Accding to the Dalton's Law the total pressure in the flask must be the sum of the pressures of the CH4 and the water vapor. The presure exerted by the water vapour and methane results molar mass of methane is higher.

5) Before start experiment we should determine the mass of the flask. The volume (weight) of the gas is the container (flask) volume. So the weight of the gas = Tolal weight of container (after collecting the gas)-Empty weight of the container.

0.07894= (0.293/M.Wt)x23.56

0.293/m.Wt= 0.0033

M.Wt =97.6.

3)

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