# Answer to Question #715 in Inorganic Chemistry for Thomasina Frimpong-LeClair

Question #715

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])

Expert's answer

Let's find the mass of sodium in the cube.The volume of cube is V = a

Thus

N = N

^{3}= 0.5375^{3}= 0.155 cm^{3}, where a - is the edge length of cube.The "volume of sodium" in this cube V_{sodium}= 0.68 * 0.155 = 0.1054 cm^{3}. 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 * 10^{23}* 0.0044 = 2.67 * 10^{21}Answers: N = 2.67* 10^{21}, m = 0.102 g.
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