Question

Which sample contains a total of 3.0 x 10^23 molecules?

Select one:
a. 80 g of Br2
b. 4.0 g of He
c. 4.0 g of H2
d. 14 g of Li

Answer 1

80 g of Br2 contains 3.0 x 10^23 molecules because its molar mass is about 160 g/mol and 80 g of Br2 is half a mole, which corresponds to half of Avogadro's number. So the correcct option is a.

Step-by-step explanation:

To determine which sample contains a total of 3.0 x 10^23 molecules, we need to use Avogadro's number, which is 6.022 x 10^23 molecules/mol. This number represents the amount of molecules in one mole of a substance. We will calculate the number of moles for each sample and see which corresponds to half a mole since half of Avogadro's number is 3.0 x 10^23 molecules.

1. For Br2 (80g), the molar mass of Br2 is about 160 g/mol (79.9 g/mol for Br), so 80 g is exactly 0.5 moles, which is the amount that contains 3.0 x 10^23 molecules.
2. For He (4.0g), the molar mass is 4 g/mol, so 4.0 g is 1 mole, which contains twice the amount we need.
3. For H2 (4.0g), the molar mass is 2 g/mol, so 4.0 g is also 2 moles, which contains four times the amount of molecules we are looking for.
4. Lastly, for Li (14g), with a molar mass of about 6.94 g/mol, 14 g corresponds to roughly 2 moles, which again is more than the required amount of molecules.

Therefore, sample a (80 g of Br2) is the correct answer as it contains 3.0 x 10^23 molecules.

Answer 2

Given that 1 mole contains 6.02x10^23 molecules, 3.0x10^23 is just around half a mole. Then we check the number of moles for each choice:

A. This is approximately half a mole, since the molar mass of Br2 is 159.8 g/mol.
B. He has a molar mass around 4 g/mol, so this is 1 mole.
C. H2 has a molar mass of 2.02 g/mol, so this is 2 moles.
D. Li has a molar mass of around 6.97 g/mol, so this is around 2 moles.

Therefore the only choice that fits is A. 80 g of Br2.