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

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Chapter7: Solutions And Colloids
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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
Transcribed Image Text: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
Transcribed Image Text: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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