Which sample contains a total of 3.0 x 10^23 molecules? a. 14 g of li b. 4.0 g of he c. 80 g of br2 d. 4.0 g of h2

Answer is: c. 80 g of Br2.
1) n(Li) = 14 g ÷ 7 g/mol
n(Li) = 2 mol.
N(Li) = 2 mol · 6·10²³ 1/mol.
N(Li) = 1,2·10²⁴.
2) n(He) = 4 g ÷ 4 g/mol.
n(He) = 1 mol.
N(He) = 1 mol · 6·10²³ 1/mol.
N(He) = 6·10²³.
3) n(Br₂) = 80 g ÷ 160 g/mol.
n(Br₂) = 0,5 mol.
N(Br₂) = 0,5 mol · 6·10²³ 1/mol.
N(Br₂) = 3·10²³.
4) n(H₂) = 4 g ÷ 2 g/mol = 2 mol.
N(H₂) = 1,2·10²⁴.

C. which metal is less dense