16 1 Model Recall the density of water is 1000 kg m3 m Solve The mass of lead 11 300 kg m 2 0 m V Pb Pb Pb 3 3 mass its volume must be 22 600 kg For water to have the same V water m water water 22 600 kg 1000 kg m 3 22 6 m 3 Assess Since the lead is 11 3 times as dense we expect the water to take 11 3 times the volume 16 2 Model Assume the nucleus is spherical Solve The volume of the uranium nucleus is V 3 R 4 3 4 3 7 5 10 15 m 1 767 10 42 3 m 3 The density of the uranium nucleus is nucleus m V nucleus nucleus 25 4 0 10 1 767 10 kg 42 m 3 17 2 3 10 kg m 3 Assess This density is extremely large compared to the typical density of materials 16 3 Solve The volume of the aluminum cube is 10 3 m3 3 2700 kg m 1 0 10 m V 3 Al Al m Al and its mass is 3 2 7 kg The volume of the copper sphere with this mass is V Cu r Cu 3 4 3 m Cu Cu 2 7 kg 8920 kg m 3 3 027 10 m 4 3 r Cu 3 3 027 10 m 4 3 4 1 3 0 042 m The diameter of the copper sphere is 0 0833 m 8 33 cm Assess The diameter of the sphere is a little less than the length of the cube and this is reasonable considering the density of copper is greater than the density of aluminum 16 4 Model The volume of a hollow sphere is V 4 p 3 3 r out 3 r in Solve We are given Solve the above equation for inr m 0 690 kg r out 0 050 m and we know that for aluminum r 3 2700 kg m r in 3 3 r out V 4 p 3 3 3 r out m r 4 p 3 3 0 050 m 3 0 040 m 0 690 kg 2700 kg m 4 p 3 3 So the inner diameter is 8 0 cm Assess We are happy that the inner diameter is less than the outer diameter and in a reasonable range 16 5 Solve The volume of the aluminum cube V 8 0 10 m 6 3 and its mass is M V 2700 kg m 8 0 10 m 0 0216 kg 21 6 g 6 3 3 One mole of aluminum 27Al has a mass of 27 g The number of atoms is N 23 6 02 10 atoms 1 mol 1 mol 27 g 21 6 g 23 4 8 10 atoms Assess This is almost one mole of atoms which is a reasonable value 16 6 Solve The volume of the copper cube is 8 0 10 6 m3 and its mass is M V 6 8920 kg m 8 0 10 m 0 07136 kg 3 3 71 36 g Because the atomic mass number of Cu is 64 one mole of Cu has a mass of 64 g The number of moles in the cube is n 1 mol 64 g 71 36 g 1 1 mol Assess This answer is in the same ballpark as the previous exercise a The number density is defined as 16 7 Solve N V where N is the number of particles occupying a volume V Because Al has a mass density of 2700 kg m3 a volume of 1 m3 has a mass of 2700 kg We also know that the molar mass of Al is 27 g mol or 0 027 kg mol So the number of moles in a mass of 2700 kg is n 2700 kg 1 mol 0 027 kg 1 00 10 mol 5 The number of Al atoms in 1 00 105 mols is N nN A 1 00 10 mol 6 02 10 atoms mol 23 5 28 6 02 10 atoms Thus the number density is N V 28 6 02 10 atoms 1 m 3 28 6 02 10 atoms m 3 b Pb has a mass of 11 300 kg in a volume of 1 m3 Since the atomic mass number of Pb is 207 the number of moles in 11 300 kg is n 11 300 kg 1 mole 0 207 kg The number of Pb atoms is thus N nNA and hence the number density is N nN V V A 11 300 kg 0 207 kg 23 6 02 10 atoms mol 1 mol 3 1 m 28 3 28 10 atoms 3 m Assess We expected to get very large numbers like this 16 8 Solve The mass density is and the mass m of each atom by M Nm M V Combining these the atomic mass is The mass M of the sample is related to the number of atoms N m M N V N N V 3 1750 kg m 28 4 39 10 atoms m 3 3 986 10 26 kg atom the atomic mass in m A u where A is the atomic mass number Thus A m 1 u 3 986 10 1 661 10 26 27 kg kg 24 The element s atomic mass number is 24 Assess This is a reasonable answer for an isotope of neon although neon is a gas at normal temperatures sodium or magnesium 4 r 3 3 16 9 Model Assume the gold is shaped into a solid sphere of volume V Visualize We want to know 1 0 mol of gold has a mass of 197 g or 0 197 kg Table 16 1 gives Solve r 2 D Because the atomic mass number of gold is 197 3 19 300 kg m V r 3 0 0135 m 1 35 cm M 3 4 3 0 197 kg 3 4 19 300 kg m 3 r 2 2 70 cm D Assess This is about the right size for a chunk that contains one mole of material 3 r M 4 3 2 1 35 cm 16 10 Solve The mass of mercury is M V 13 600 kg m 10 cm 3 3 0 136 kg 136 g 3 6 10 m 3 1 cm and the number of moles is n M M mol 0 136 g 201 g mol 0 6766 mol The mass of aluminum with 0 6766 mol of Al is M 0 6766 mol M mol 0 6766 mol 18 27 g 0 01827 kg 27 g mol This mass M of aluminum corresponds to a volume of V M 0 01827 kg 3 2700 kg m 3 6 8 10 m 6 8 cm 6 3 Assess We expected an answer in the same order of magnitude The size of atoms doesn t vary as much as the density of atoms from element to element 16 11 Solve The lowest temperature is In the same way the highest temperature is …
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