Interpretation Introduction To determine: Mass of sodium chloride. Explanation Given: Freezing point of solution ( T solution ) is − 10 ° C . Volume of water ( V ) is 1 L . Density of water ( ρ ) is 1 g / mL . Explanation: 1 L is equal to 1000 mL. Volume of water ( V ) is 1000 mL The density of a solution is equal to the ratio of mass and volume of solution. Here, mass of water can be calculated as m w = ρ × V Here, m w is the mass of water, ρ is the density of water, and V is the volume of water. Substitute 1 g / mL for ρ and 1000 mL for V . m w = 1 g / mL × 1000 mL = 1000 g Mass of water is 1000 g . 1 g is equal to 10 − 3 kg. Convert grams to kilograms as follows: 1000 g = 1000 g × ( 10 − 3 kg 1 g ) = 1 kg Mass of water in kilograms is 1 kg . Freezing point of water ( T pure ) is 0 ° C . Calculate the molality of the solution as follows. Use the formula for depression in freezing point for a strong electrolyte: Δ T f = i × K f × m Δ T f is the change in the temperature of freezing point, K f is the freezing point depression constant, m is the molality, and i is the van’t Hoff factor. For water, K f = 1.86 ° C / m . Δ T f = T pure − T solution Substitute 0 ° C for T pure and − 10 ° C for T solution : Δ T f = 0 ° C − ( − 10 ° C ) = 10 ° C Substitute 1.86 ° C / m for K f , 2 for i , and 10 ° C for Δ T f : 10 ° C = 2 × 1

Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for Chemistry: A Molecular Approach (4th Edition)

4th Edition

ISBN: 9780134162430

Author: Nivaldo J. Tro

Publisher: PEARSON

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Chapter 13, Problem 89E

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To determine: Mass of sodium chloride.

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Chapter 13, Problem 88E

Chapter 13, Problem 90E

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Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for Chemistry: A Molecular Approach (4th Edition)

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