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STARTING AMOUNT X Glycerol (CH:O) is a byproduct of soapmaking. What is the molality of a solution made by dissolving 155 g C.H.Os per L water ? The density of water is 1 kg water/L. ADD FACTOR x( ) 1680 g C₂H₂O₂/mol 6.022 x 1023 16.8 m C₂H₂O₂ 1 DELETE H 1.68 1000 0.594 ANSWER 15.5 594 mL water g C₁H.O/L water mol CsH₂Os kg water/L 59.4 g C₂H₂O₂ 0.001 RESET J 92.09 kg water 155 L water

STARTING AMOUNT X Glycerol (CH:O) is a byproduct of soapmaking. What is the molality of a solution made by dissolving 155 g C.H.Os per L water ? The density of water is 1 kg water/L. ADD FACTOR x( ) 1680 g C₂H₂O₂/mol 6.022 x 1023 16.8 m C₂H₂O₂ 1 DELETE H 1.68 1000 0.594 ANSWER 15.5 594 mL water g C₁H.O/L water mol CsH₂Os kg water/L 59.4 g C₂H₂O₂ 0.001 RESET J 92.09 kg water 155 L water

Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:John W. Moore, Conrad L. Stanitski
Chapter13: The Chemistry Of Solutes And Solutions
Section: Chapter Questions
Problem 127QRT
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STARTING AMOUNT X = Glycerol (C3H8O3) is a byproduct of soapmaking. What is the molality of a solution made by dissolving 155 g C3H8O3 per L water ? The density of water is 1 kg water/L ADD FACTOR x( ) 1680 6.022 x 1023 16.8 1 DELETE t 1.68 1000 0.594 ANSWER 15.5 594 59.4 g C3H₂O3/mol m C3HO3 mL water g C3H8O/L water g C3HsO3 mol C3H₂O₂ kg water/L 0.001 RESET 3 92.09 kg water 155 L water
Transcribed Image Text:STARTING AMOUNT X = Glycerol (C3H8O3) is a byproduct of soapmaking. What is the molality of a solution made by dissolving 155 g C3H8O3 per L water ? The density of water is 1 kg water/L ADD FACTOR x( ) 1680 6.022 x 1023 16.8 1 DELETE t 1.68 1000 0.594 ANSWER 15.5 594 59.4 g C3H₂O3/mol m C3HO3 mL water g C3H8O/L water g C3HsO3 mol C3H₂O₂ kg water/L 0.001 RESET 3 92.09 kg water 155 L water
Expert Solution
Step 1

The molality of the solution is given by the formula,

Molality =moles of the solutemass of the solvent in kg

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