Question

1. we obtain uranium - 235 from u - 238 by fluorinating the 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.

Explanation:

Step1: Calculate the molar mass of \(^{235}UF_6\)

The molar mass of \(F\) is \(19.00\space g/mol\).
For \(^{235}UF_6\), \(M_{235}=235.054 + 6\times19.00=235.054 + 114=349.054\space g/mol\)

Step2: Calculate the molar mass of \(^{238}UF_6\)

For \(^{238}UF_6\), \(M_{238}=238.051+6\times19.00 = 238.051 + 114=352.051\space g/mol\)

Step3: Apply Graham's law of effusion

Graham's law is \(\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}\), where \(r\) is the effusion rate and \(M\) is the molar mass.
Let \(r_{238}\) be the effusion rate of \(^{238}UF_6\) and \(r_{235}\) be the effusion rate of \(^{235}UF_6\). Then \(\frac{r_{238}}{r_{235}}=\sqrt{\frac{M_{235}}{M_{238}}}\)
Substitute \(M_{235} = 349.054\space g/mol\) and \(M_{238}=352.051\space g/mol\)
\(\frac{r_{238}}{r_{235}}=\sqrt{\frac{349.054}{352.051}}\)
\(\frac{r_{238}}{r_{235}}=\sqrt{0.9915}\approx0.9957\)

Answer:

The ratio of effusion rates \(\frac{r_{238}}{r_{235}}\approx0.9957\)