Question

Question #f833d

Answer 1

`"235.091 u"`

Explanation:

All you have to do to solve this problem is use the fact that the atomic mass of uranium depends on the isotopic masses of its isotopes and on their respective abundances.

`"atomic mass" = sum_(i) "isotope mass"_i * "abundance"_i`

You know that you're dealing with three isotopes, uranium-234, uranium-235, and uranium-238, for which you know

`""^234"U" -> "234.041 u", 0.0057%`

`""^235"U" -> x, 0.72%`

`""^238"U" -> "238.051 u", 99.27%`

This means that you an set up your equation like this - I'll use decimal abundances, which are simply percent abundances divided by `100`.

`0.000057 * 234.041 + 0.0072 * x + 0.9927 * 238.051 = 238.019`

All you have to do now is solve for `x`, the isotopic mass of uranium-235.

`x = (238.019 - 0.0133403 - 236.313)/0.0072`

`x = 1.692658/0.0072 = color(green)("235.091 u")`