# Question efe8d

Jan 14, 2018

$2.6 \cdot {10}^{24}$

#### Explanation:

The idea here is that you can convert the mass of uranium-235 to moles and then to atoms by using the isotope's molar mass and Avogadro's constant.

Since the problem doesn't provide you with the actual molar mass of uranium-235, you can approximate it to be

M_ ("M" quad ""^235"U") ~~ "235 g mol"^(-1)

So, you know that $1$ mole of uranium-235 will have a mass of $\text{235 g}$. You also know that $1$ mole of uranium-235 will contain $6.022 \cdot {10}^{23}$ atoms of uranium-235, as given by Avogadro's constant.

This means that you have

"1 mole" quad ""^235"U" = {(6.022 * 10^(23) quad "atoms of" quad ""^235"U" -> color(blue)("from Avogadro's constant")), ("235 g" -> color(blue)("from the molar mass of" quad ""^235"U")) :}#

You can thus say that your sample will contain

$1.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{kg"))) * (10^3 quad "g")/(1color(red)(cancel(color(black)("kg")))) * (6.022 * 10^(23)quad "atoms of" quad ""^235 "U")/(235color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)(2.6 * 10^(24) quad "atoms of" quad ""^235"U}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the mass of uranium-235.