Apply algebraic properties, formulas, and relationships to perform operations on real-world problems (e.g., solve for: molality.• Molality is the number of moles of solute dissolved in a kilogram of solvent.molality = moles of solute or molal, mkg of solventSample Problem #1: What is the molality of the example sucrose solution?First Step: Known and Unknown Quantitiesknownunknownmass of sucrose = 0.030464 g C12H2201molality C12H2201 = ? mmass of water = 0.981256 g H,0Second Step: Formulamolality of C12H22O11 = mol of C12H22O11 kg of H,0Third Step: Rearranging, intermediate calculations, and unit conversionsNo rearranging required.Intermediate calculations:molar mass of C12H2201 = (12 x 12.011 g) + (22 x 1.0079 g) + (11 x 15.9994 g) = 342.299 g# mol C12H2201 = 0.030464 g C12H,O1 x1 mol C,,H„01 = 0.000089 mol C12H22O11342.299 g C12H2201Unit conversions:0.981256 g H,Ox_1.00000 kg_ = 0.000981256 kg H,011000.00 gFourth Step:DELL「田 Fourth Step:molality, C12H2201 = 0.000089 mol C1,HO1 = 0.091 m0.000981256 kg H20Sample Problem #2: What is the molality of a solution of 0.20 g of benzoic acid, C,H,O2, in 20.0 g of naphthalene?moles of C,HO2 =0.20 g122 g/mol= 0.0016 mol C,H¢O2molality of C,HO2 =0.0016 mol C,H;O>0.020 kg of naphthalene= 0.080 molal (mol/kg)Note: The "molal" unit is mol of solute/kg of solvent. The molal unit is abbreviated with the lower case letter, m.The molality of the C;HgO2 solution is 0.080 m.Practice Problem #1: What is the molality of a solution of 0.50 g of glucose, CH12O6, 14.3 g of water?O 35 mO 0.35 mO 0.19 mO 0.00019 mDELL&司%24#3

Practice Problem#1: Molality is defined as the number of solute per unit mass of the solvent in…

Answered: Practice Problem #1: What is the…

Practice Problem #1: What is the molality of a solution of 0.50 g of glucose, CgH12Os, 14.3 g of water? O 35 m O 0.35 m O 0.19 m O 0.00019 m

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Concept explainers

Solutions

Solutions can be considered as homogeneous mixtures which contain two or more than two chemical substances. The chemical substance which is the greater part is defined as solvent. And the chemical substance that is dissolved in a solvent is referred to as solute. Solution chemistry plays an important role in chemistry since it supports many commercial applications of solutions.

Question

Apply algebraic properties, formulas, and relationships to perform operations on real-world problems (e.g., solve for: molality.
• Molality is the number of moles of solute dissolved in a kilogram of solvent.
molality = moles of solute or molal, m
kg of solvent
Sample Problem #1: What is the molality of the example sucrose solution?
First Step: Known and Unknown Quantities
known
unknown
mass of sucrose = 0.030464 g C12H2201
molality C12H2201 = ? m
mass of water = 0.981256 g H,0
Second Step: Formula
molality of C12H22O11 = mol of C12H22O1
1 kg of H,0
Third Step: Rearranging, intermediate calculations, and unit conversions
No rearranging required.
Intermediate calculations:
molar mass of C12H2201 = (12 x 12.011 g) + (22 x 1.0079 g) + (11 x 15.9994 g) = 342.299 g
# mol C12H2201 = 0.030464 g C12H,O1 x
1 mol C,,H„01 = 0.000089 mol C12H22O1
1
342.299 g C12H2201
Unit conversions:
0.981256 g H,Ox_1.00000 kg_ = 0.000981256 kg H,0
1
1000.00 g
Fourth Step:
DELL
「田

Fourth Step:
molality, C12H2201 = 0.000089 mol C1,HO1 = 0.091 m
0.000981256 kg H20
Sample Problem #2: What is the molality of a solution of 0.20 g of benzoic acid, C,H,O2, in 20.0 g of naphthalene?
moles of C,HO2 =
0.20 g
122 g/mol
= 0.0016 mol C,H¢O2
molality of C,HO2 =
0.0016 mol C,H;O>
0.020 kg of naphthalene
= 0.080 molal (mol/kg)
Note: The "molal" unit is mol of solute/kg of solvent. The molal unit is abbreviated with the lower case letter, m.
The molality of the C;HgO2 solution is 0.080 m.
Practice Problem #1: What is the molality of a solution of 0.50 g of glucose, CH12O6, 14.3 g of water?
O 35 m
O 0.35 m
O 0.19 m
O 0.00019 m
DELL
&
司
%24
#3

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