We separate U-235 from U-238 by fluorinating a sample of urani...

Question

We separate U-235 from U-238 by fluorinating a sample of uranium to form UF₆ (which is a gas) and then taking advantage of the different rates of effusion and diffusion for compounds containing the two isotopes. Calculate the ratio of effusion rates for ²³⁸UF₆ and ²³⁵UF₆. The atomic mass of U-235 is 235.054 amu and that of U-238 is 238.051 amu.

Accepted Answer

Identify the formula for effusion rates: Graham's law of effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, this is expressed as $$ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} $$, where $$ r_1 $$ and $$ r_2 $$ are the effusion rates of gases 1 and 2, and $$ M_1 $$ and $$ M_2 $$ are their respective molar masses.. Determine the molar masses of $$ ^{238}	ext{UF}_6 $$ and $$ ^{235}	ext{UF}_6 $$: The molar mass of $$ ^{238}	ext{UF}_6  is $$the sum of the atomic mass of $$ ^{238}	ext{U} $$ and six times the atomic mass of fluorine ($$ 18.998 	ext{ amu} ). $$Similarly, calculate the molar mass for $$ ^{235}	ext{UF}_6 .. $$Calculate the molar mass of $$ ^{238}	ext{UF}_6 $$: $$ M_{238} = 238.051 + 6 	imes 18.998 .. $$Calculate the molar mass of $$ ^{235}	ext{UF}_6 $$: $$ M_{235} = 235.054 + 6 	imes 18.998 .. $$Apply Graham's law to find the ratio of effusion rates: Substitute the molar masses into the formula $$ \frac{r_{238}}{r_{235}} = \sqrt{\frac{M_{235}}{M_{238}}}  to $$find the ratio of effusion rates for $$ ^{238}	ext{UF}_6 $$ and $$ ^{235}	ext{UF}_6 .$$

Risposta video verificata per un problema simile:

Concetti chiave

Effusion and Diffusion

Graham's Law of Effusion

Isotopes and Atomic Mass