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

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Chapter1: Chemical Foundations
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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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