Effusion molar mass

At a given temperature and pressure, die rate of effusion of a gas, in moles per unit time, is inversely proportional to the square root of its molar mass. 

Graham s law tells us qualitatively that light molecules effuse more rapidly than heavier ones (Figure 5.8, p. 119). In quantitative form, it allows us to determine molar masses of gases (Example 5.11). 

In an effusion experiment, argon gas is allowed to expand through a tiny opening into an evacuated flask of volume 120 mL for 32.0 s, at which point the pressure in the flask is found to be 12.5 mm Hg. This experiment is repeated with a gas X of unknown molar mass at the same T and P. It is found that the pressure in the flask builds up to 12.5 mm Hg after 48.0 s. Calculate the molar mass of X. 

Effusion of gases. A gas with a higher molar mass (red molecules) effuses into a vacuum more slowly than a gas with a lower molar mass (gray molecules). 

Reality Check Because it takes longer for gas X to effuse, it must have a larger molar mass than argon. It does ... 

Use Graham s law to relate rate of effusion to molar mass. 

At 25°C and 380 mm Hg, the density of sulfur dioxide is 1.31 g/L. The rate of effusion of sulfur dioxide through an orifice is 4.48 mL/s. What is the density of a sample of gas that effuses through an identical orifice at the rate of 6.78 mL/s under the same conditions What is the molar mass of the gas ... 

This relation can be used to estimate the molar mass of a substance by comparing the time required for a given amount of the unknown substance to effuse with that required for the same amount of a substance with a known molar mass. 

The same volume of vapor of a volatile compound extracted front Caribbean sponges takes 120. s to effuse through the same barrier under the same conditions. What is the molar mass of this compound ... 

We have seen that effusion reveals that the average speed of molecules in a gas is inversely proportional to the square root of their molar mass. In effusion experiments at different temperatures, we find that the rate of effusion increases as the temperature is raised. Specifically, for a given gas, the rate of effusion increases as the square root of the temperature ... 

What is the molar mass of a compound that takes 2.7 times as long to effuse through a porous plug as it did for the same amount of XeF, at the same temperature and pressure ... 

A hydrocarbon of empirical formula C,H, takes 349 s to effuse through a porous plug under the same conditions of temperature and pressure, it took 210. s for the same number of molecules of argon to effuse. What is the molar mass and molecular formula of the hydrocarbon ... 

The enrichment procedure uses the small mass difference between the hexafluorides of uranium-235 and uranium-238 to separate them. The first procedure to be developed converts the uranium into uranium hexafluoride, UFfl, which can be vaporized readily. The different effusion rates of the two isotopic fluorides are then used to separate them. From Graham s law of effusion (rare of effusion l/(molar mass)1/2 Section 4.9), the rates of effusion of 235UFfe (molar mass, 349.0 g-mol ) and 238UF6 (molar mass, 352.1 g-mol ) should be in the ratio... 

Rates of molecular motion are directly proportional to molecular speeds, so Equation predicts that for any gas, rates of effusion and diffusion increase with the square root of the temperature in kelvins. Also, at any particular temperature, effusion and diffusion are faster for molecules with small molar masses. 

B The two rates of effusion are related as the square root of the ratio of the molar masses of the two gases. The lighter gas, H2, effuses faster, and thus requires a shorter time for the same amount of gas to effuse. 

Effusion time is proportional to the square root of molar mass. 

We can use Graham s law to determine the rate of effusion of an unknown gas knowing the rate of a known one or we can use it to determine the molecular mass of an unknown gas. For example, suppose you wanted to find the molar mass of an unknown gas. You measure its rate of effusion versus a known gas, H2. The rate of hydrogen effusion was 3.728 mL/s, while the rate of the unknown gas was 1.000 mL/s. The molar mass of H2 is 2.016 g/mol. Substituting into the Graham s law equation gives ... 

B—Lighter gases effuse faster. The only gas among the choices that is lighter than methane is helium. To calculate the molar mass, you would begin with the molar mass of methane and divide by the rate difference squared ... 

Urrre= root-mean-square speed KE= kinetic energy r = rate of effusion M= molar mass jc= osmotic pressure /= van t Hoff factor... 

H2 because hydrogen is a diatomic element. Consult your periodic table (or your memory, if you re that good) to obtain the molar masses of hydrogen gas (2.02 g/mol) and neon gas (20.18 g/mol). Finally, plug those values into the appropriate places within Grahcim s law, and you can see the ratio of effusion speed. 

Hydrogen, H2. The question states that the ratio of the rates is 4.0. Oxygen gas is a diatomic element, so it s written as O2 and has a molar mass of 32.00 g/mol. Substitute these known values into Graham s law to determine the molar mass of the unknown gas. The problem states that the unknown gas effuses at a rate 4.0 times faster than oxygen, so put the unknown gas over oxygen for the ratio. (In short, the unknown gas is A, and the oxygen is B). 

Thomas Graham, a nineteenth-century Scottish chemist, did a series of experiments on the rate of effusion of gases. He found that at constant temperature, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. This observation is now known as Graham s law of effusion. It follows that for two gases A and B ... 

FIGURE 4.21 In effusion, the molecules of one substance escape through a small hole in a barrier. In both effusion and diffusion, the rate increases with increasing temperature and decreases with increasing molar mass. 

In an experiment to determine the molar mass of a newly synthesized chlo-rofluorocarbon (CFC) gas for use in a refrigeration system, it was found that. 25 mL of the gas effused through a porous barrier in 65 s. The same volume... 

STRATEGY Rate of effusion is inversely proportional to the time taken for a I given volume to effuse the faster the effusion, the less time it takes for a given volume to effuse. We combine that idea with Graham s law to obtain, the relation between the effusion times and the molar masses ... 

Molar mass of CFC /time for CFC to effuse 2 Molar mass of Ar time for Ar to effuse /... 

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