### Homework problem E3

Posted: **Wed Oct 02, 2019 1:20 am**

by **Doris Cho 1D**

Not really sure where to even start for the problem, but would you use the formula n=m/M and the m would be = to the 70 g.mol-1 and replace the M with the atomic mass of gallium, then multiply by Avogadro's number to the find the number of moles and then do something similar for finding the amount of astatine atoms?

### Re: Homework problem E3 [ENDORSED]

Posted: **Wed Oct 02, 2019 1:29 am**

by **Charisse Vu 1H**

I believe this question is simpler than what it seems. In the question, the gallium atoms are literally represented by the blue circles on the scale. If you count them, you would see that there are 9 "atoms" of gallium on the left hand side of the scale. Comparing the molar masses of gallium and astatine, you would see that the molar mass of astatine is 3 times more than that of gallium's. You don't necessarily need to use Avogadro's constant because the scale acts much like an equal sign and multiplying the constant to both sides would essentially do nothing. So, since the molar mass of astatine is 3 times more than that of gallium's you would simply need to divide the number of gallium atoms by 3 to determine the number of astatine atoms needed to have the masses be equal.