Answer to Question #715 in Inorganic Chemistry for Thomasina

Question

Sodium metal is so soft that it can be cut with a knife. You cut a perfect cube of sodium with an edge length of 0.5375 cm. If 68% of the cube’s volume contains sodium atoms, how many sodium atoms are in this cube? What is the mass, in grams, of sodium present in the cube? (density of sodium = 0.97 g/cm[sup]3[/sup])

Answer 1

Let's find the mass of sodium in the cube.The volume of cube is V = a³ = 0.5375³ = 0.155 cm³, where a - is the edge length of cube.The "volume of sodium" in this cube V_sodium = 0.68 * 0.155 = 0.1054 cm³. m = ρ * V_sodium = 0.97 * 0.105 = 0.102 g.μ = m/M = 0.102 / 23 = 0.0044 mol. (atomic weight of Na = 23 g/mol)

Thus
N = N_A * μ = 6.022 * 10²³ * 0.0044 = 2.67 * 10²¹Answers: N = 2.67* 10²¹, m = 0.102 g.