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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&rsquo;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])
1
2010-10-05T19:37:47-0400
Let&#039;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 &quot;volume of sodium&quot; in this cube Vsodium = 0.68 * 0.155 = 0.1054 cm3. m = &rho; * Vsodium = 0.97 * 0.105 = 0.102 g.&mu; = m/M = 0.102 / 23 = 0.0044 mol. (atomic weight of Na = 23 g/mol)

Thus
N = NA * &mu; = 6.022 * 1023 * 0.0044 = 2.67 * 1021Answers: N = 2.67* 1021, m = 0.102 g.

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