Explanation Given info: The spacing of adjacent atoms in a crystal of sodium chloride is 0.282 nm and the mass of sodium chloride is 3.82 × 10 − 26 kg and the mass of the chlorine atom is 5.89 × 10 − 26 kg . Write the expression for volume of an atom occupies a cube. V = ( a ) 3 (1) Here, V is the volume of the atom a is the side length of the cube where atom occupies Substitute 0.282nm for a in equation (1)to find V . V = ( 0.282 nm × 1 m 10 9 nm ) 3 = ( 0.282 × 10 − 9 m ) 3 = 2.24 × 10 − 29 m 3 The NaCl contains equal number of Na and Cl atoms and the average mass of atoms in NaCl crystal is given by m = 1 2 ( m N a + m C l ) (2) Here, m is the average mass of atoms in NaCl crystal m N a is the mass of sodium atom m C l is the mass of the chlorine atom Substitute 3.82 × 10 − 26 kg for m N a and 5.89 × 10 − 26 kg for m C l in equation (2) to find m

13th Edition

ISBN: 9781269542661

Author: YOUNG

Publisher: PEARSON C

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Chapter 42, Problem 42.15E

To determine

The density of sodium chloride.

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Chapter 42, Problem 42.14E

Chapter 42, Problem 42.16E

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Chapter 42 Solutions

MASTERINGPHYSICS W/ETEXT ACCESS CODE 6

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