Fundamental Exercises E3 [ENDORSED]

[Keshav Bhatnagar 1H]

Please explain problem E3 in the fundamental exercises.

[Venya Vaddi 1L]

I am not sure if you are referring to problem E3 in the 6th or 7th edition, but here is an explanation for E3 in the 7th edition.

Based on the given molar masses, 1 mole of gallium atoms is 70 grams, and 1 mole of astatine atoms are 210 grams. In order for the two pans to be equal, the masses on each pan must be the same. Thus, both pans must be equal to 210 grams. Since gallium is 70 g/mol, you would need three times as many gallium atoms as you have astatine atoms. One mole of astatine has 6.022 x 10²³ atoms. Since you need three times the number of gallium atoms, you need 3 x (6.022 x 10²³) atoms of gallium.

[Matthew Tran 1H]

I was initially confused by this question at first, but you actually have to count the number of gallium atoms on the left side of the balance (9). Then you could solve. Since astatine has a molar mass that is three times larger than that of gallium, the number of astatine atoms on the right side would need to be 1/3 the number of gallium atoms = 3 in order to balance the sides.

[Jonathan Omens 1K]

I agree with Matthew. Because the question says that the lab can manipulate individual atoms, it is safe to assume that each circle on the scale represents a single atom. Therefore, the answer will be 3 atoms of astatine rather than 3 moles of astatine.