During the discussion of gaseous diffusion for enriching uranium, it was claimed that \(^{235} \mathrm{UF}_{6}\) diffuses \(0.4 \%\) faster than \(^{238} \mathrm{UF}_{6}\). Show the calculation that supports this value. The molar mass of \(^{235} \mathrm{UF}_{6}=235.043930+6 \times\) \(18.998403=349.034348 \mathrm{g} / \mathrm{mol},\) and the molar mass of \(^{238} \mathrm{UF}_{6}=238.050788+6 \times 18.998403=352.041206 \mathrm{g} /\)mol.

Graham's Law of Effusion is a valuable tool in predicting how different gases will behave under similar conditions. To clarify the concept of effusion rate calculation, let us take a closer look at the phenomenon. Effusion is the process by which gas particles pass through a tiny opening from one chamber to another. It differs from diffusion as the latter involves gases spreading out throughout a given space.

According to Graham's Law, the effusion rate is inversely proportional to the square root of the molar mass of the gas. This can be represented mathematically by the equation: \[\frac{Rate_1}{Rate_2} = \frac{Rate(\text{Gas 1})}{Rate(\text{Gas 2})} = \sqrt{\frac{Molar Mass_2}{Molar Mass_1}}\]
Here, \(Rate_1\) and \(Rate_2\) are the effusion rates of Gas 1 and Gas 2, respectively. For instance, if you have two isotopes of the same compound, the lighter isotope will effuse at a faster rate than the heavier isotope due to its lesser molar mass. Understanding and applying this calculation is not only essential in solving homework problems but also is fundamental in fields like chemical manufacturing and isotopic enrichment processes.

Separation of isotopes is a significant application of Graham's Law of Effusion, predominantly illustrated in the context of uranium enrichment. Enrichment involves increasing the concentration of the lighter, fissionable uranium-235 isotope relative to the heavier uranium-238 isotope. According to Graham's Law, lighter isotopes effuse more quickly, providing a means to separate them. In the context of gaseous diffusion for uranium enrichment:

\[\[\begin{equation}\frac{Rate(^{235} UF_6)}{Rate(^{238} UF_6)} = \sqrt{ \frac{Molar Mass(^{238} UF_6)}{Molar Mass(^{235} UF_6)}}\end{equation}\]\]
By using this calculated ratio, one can design a system where the slightly faster effusion rate of the lighter isotope (\(^{235} UF_6\)) leads to its gradual accumulation. This principle of separation has massive implications for nuclear power generation and the creation of nuclear medicine isotopes. It's a delicate and continuous process, requiring numerous stages of repeating this separation to achieve the desired level of enrichment.

The molar mass is a fundamental concept in chemistry that influences numerous physical properties of substances, including effusion rate as described by Graham's Law. Specifically, molar mass is the mass in grams of one mole of molecules or atoms of a substance. It can be found by summing the atomic masses of all atoms in the molecule.

In the exercise, the molar mass of uranium hexafluoride isotopes was calculated by adding the mass of the uranium isotope to six times the atomic mass of fluorine. The molar masses of \(^{235} UF_6\) and \(^{238} UF_6\) are central to the problem as they serve as the inputs for Graham's Law when calculating effusion rates:\[Molar Mass(^{235} UF_6) = 235.043930 + 6 \times 18.998403 = 349.034348 g/mol\]\[Molar Mass(^{238} UF_6) = 238.050788 + 6 \times 18.998403 = 352.041206 g/mol\]
The intriguing part about the molar mass is its close relationship with the periodic table and stoichiometry. It acts as a bridge between the microscopic particles and the macroscopic measurements, allowing chemists to calculate the quantities needed for reactions and analyze the outcomes using scales and other lab equipment.