Final answer:
The sample containing 3.0 x 10²³ molecules is 80 g of bromine (Br²), which equates to 0.5 moles, exactly half of Avogadro's number of molecules. Therefore correct option is A
Step-by-step explanation:
The question "Which sample contains a total of 3.0 x 10²³ molecules?" relates to understanding Avogadro's number (6.02 x 10²³), molecular mass, and molar mass in chemistry.
To determine which sample contains 3.0 x 10²³ molecules, we have to calculate the moles in each sample and see which one is equal to 0.5 moles because 0.5 moles of a substance contains half of Avogadro's number of molecules.
For bromine (Br₂), with a molar mass of about 160 g/mol, 80 g corresponds to 0.5 moles, which means it has 3.0 x 10²³ molecules.
Helium (He) has a molar mass of 4 g/mol, so 4.0 g has 1 mole, which is twice the required number of molecules. Hydrogen gas (H₂) has a molar mass of 2 g/mol, so 4.0 g would constitute 2 moles, again too much. Lithium (Li) has a molar mass of about 7 g/mol, which makes 14 g equal to 2 moles, too many molecules.
Therefore, the correct answer is 80 g of Br².