Explanation Given: Volume of liquid nitrogen, V = 1 .0 L Volume of the closet = 1.0 m × 1.0 m × 2.0 m Temperature, T = 25 ° C Pressure, P = 1.0 atm Density of liquid nitrogen, d = 0 .807 g/mL Explanation: V o l u m e o f t h e c l o s e t = 1.0 m × 1.0 m × 2.0 m = 2.0 m 3 As density is given in g/mL , the volume of the closet is converted into mL . 1 m 3 = 1000 L ∴ 2 .0 m 3 = 1000 L × 2 .0 m 3 1 m 3 = 2000 L Thus, the volume of the closet is 2000 L . As the closet is filled with air and liquid nitrogen, the volume of the closet is equal to the volume of the air and volume of nitrogen. However, the volume of nitrogen is 1.0 L ; therefore, the volume of air present in the closet is 1999 L . In order to determine the amount of gaseous nitrogen that would be formed when the liquid nitrogen evaporates, the mass should be determined. The mass of liquid nitrogen can be determined from its given density. Therefore, Mass of nitrogen = Density of liquid nitrogen × Volume of liquid nitrogen To find out the mass of liquid nitrogen, convert the density into g/L as follows: Density in g/L = 0 .807 g mL × 1000 mL L [ ∵ 1 L = 1000 mL ] = 807 g/L Substitute density of liquid nitrogen = 807 g/L and volume of the liquid nitrogen = 1.0 L in the above expression to get Mass of liquid nitrogen = 807 g L × 1.0 L = 807 g Molar mass of liquid nitrogen = 2 × 14 = 28 g / m o l Number of moles ( n ) = Mass of N 2 Molar mass of N 2 = 807 g 28 g /mol = 28

Chemistry: A Molecular Approach (4th Edition)

4th Edition

ISBN: 9780134112831

Author: Nivaldo J. Tro

Publisher: PEARSON

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Chapter 5, Problem 46E

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To determine: Percentage of nitrogen gas displaced from the container.

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Chapter 5, Problem 45E

Chapter 5, Problem 47E

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Chemistry: A Molecular Approach (4th Edition)

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