Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
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Chapter 11, Problem 11.66QP
Interpretation Introduction

Interpretation: The table representing the different ions in sea water is given. The molality of each ion is to be calculated.

Concept introduction: Molality is defined as the number of moles of solute that is present in 1kg of solvent

To determine: The molality of each given ion is to be calculated.

Expert Solution & Answer
Check Mark

Answer to Problem 11.66QP

Solution

The molarity for given ions in terms of molality is tabulated as,

Ions Molality (mol/kg)
Na+ 0.4753
K+ 1.02×10-2
Mg2+ 5.3×10-2
Ca2+ 1.03×10-2
Sr2+ 9.08×10-5
Cl 5.5×10-1
SO42 2.83×10-2
HCO3 2.06×10-3
Br 8.5×10-4
B(OH)3 4.2×10-4
F 6.9×10-5

Explanation of Solution

Explanation

Given

The density for sea water is 1.022g/mL

The table for molarity is given as,

Ions Concentration (mmol/L)
Na+ 480.57
K+ 10.46
Mg2+ 54.14
Ca2+ 10.53
Sr2+ 0.0928
Cl 559.40
SO42 28.93
HCO3 2.11
Br 0.865
B(OH)3 0.426
F 0.070

Table 1

The conversion of mmol to mol is done as,

1mmol=0.001mol

Hence, the conversion of 480.57mmol to mol is done as,

480.57mmol=480.57×0.001mol=480.57×103mol

Similarly, the conversion of 10.46mmol to mol is done as,

10.46mmol=10.46×0.001mol=10.46×103mol

Similarly, the conversion of 54.14mmol to mol is done as,

54.14mmol=54.14×0.001mol=54.14×103mol

Similarly, the conversion of 10.53mmol to mol is done as,

10.53mmol=10.53×0.001mol=10.53×103mol

Similarly, the conversion of 0.0928mmol to mol is done as,

0.0928mmol=0.0928×0.001mol=0.0928×103mol

Similarly, the conversion of 559.40mmol to mol is done as,

559.40mmol=559.40×0.001mol=559.40×103mol

Similarly, the conversion of 28.93mmol to mol is done as,

28.93mmol=28.93×0.001mol=28.93×103mol

Similarly, the conversion of 2.11mmol to mol is done as,

2.11mmol=2.11×0.001mol=2.11×103mol

Similarly, the conversion of 0.865mmol to mol is done as,

0.865mmol=0.865×0.001mol=0.865×103mol

Similarly, the conversion of 0.426mmol to mol is done as,

0.426mmol=0.426×0.001mol=0.426×103mol

Similarly, the conversion of 0.070mmol to mol is done as,

0.070mmol=0.070×0.001mol=0.070×103mol

The concentration of ions is tabulated as,

Ions Concentration (mol/L)
Na+ 480.57×103
K+ 10.46×103
Mg2+ 54.14×103
Ca2+ 10.53×103
Sr2+ 0.0928×103
Cl 559.40×103
SO42 28.93×103
HCO3 2.11×103
Br 0.865×103
B(OH)3 0.426×103
F 0.070×103

Table 2

The molar mass of Na+ is 22.99g/mol .

The molar mass of K+ is 39.0983g/mol .

The molar mass of Mg2+ is 24.30g/mol .

The molar mass of Ca2+ is 40.078g/mol .

The molar mass of Sr2+ is 87.62g/mol .

The molar mass of Cl is 35.453g/mol .

The molar mass of SO42 is 96.06g/mol .

The molar mass of HCO3 is 61.0168g/mol .

The molar mass of Br is 79.904g/mol .

The molar mass of B(OH)3 is 61.831g/mol .

The molar mass of F is 19.0g/mol .

Number of moles of a substance is given as,

Number of moles=Given massMolar mass (1)

Molarity of a solution is given as,

Molarity=Number of moles of soluteVolume of solution in litres (2)

Substitute number of moles by mass in equation (2),

Molarity=Given massMolar massVolume of solution in litres (3)

Substitute molar mass of Na+ , molarity and volume of solution in equation (3) as,

Molarity=Given massMolar massVolume of solution in litres480.57×103=Given mass22.99gmol1.0LGiven mass=11.048g

Substitute molar mass of K+ , molarity and volume of solution in equation (3) as,

Molarity=Given massMolar massVolume of solution in litres10.46×103=Given mass39.0983gmol1.0LGiven mass=0.4090g

Substitute molar mass of Mg2+ , molarity and volume of solution in equation (3) as,

Molarity=Given massMolar massVolume of solution in litres54.14×103=Given mass24.30gmol1.0LGiven mass=1.316g

Substitute molar mass of Ca2+ , molarity and volume of solution in equation (3) as,

Molarity=Given massMolar massVolume of solution in litres10.53×103=Given mass40.078gmol1.0LGiven mass=0.4220g

Substitute molar mass of Sr2+ , molarity and volume of solution in equation (3) as,

Molarity=Given massMolar massVolume of solution in litres0.0928×103=Given mass87.62gmol1.0LGiven mass=0.0081g

Substitute molar mass of Cl , molarity and volume of solution in equation (3) as,

Molarity=Given massMolar massVolume of solution in litres559.40×103=Given mass35.453gmol1.0LGiven mass=19.832g

Substitute molar mass of SO42 , molarity and volume of solution in equation (3) as,

Molarity=Given massMolar massVolume of solution in litres28.93×103=Given mass96.06gmol1.0LGiven mass=2.779g

Substitute molar mass of HCO3 , molarity and volume of solution in equation (3) as,

Molarity=Given massMolar massVolume of solution in litres2.11×103=Given mass61.0168gmol1.0LGiven mass=0.129g

Substitute molar mass of Br , molarity and volume of solution in equation (3) as,

Molarity=Given massMolar massVolume of solution in litres0.865×103=Given mass79.904gmol1.0LGiven mass=0.0691g

Substitute molar mass of B(OH)3 , molarity and volume of solution in equation (3) as,

Molarity=Given massMolar massVolume of solution in litres0.426×103=Given mass61.831gmol1.0LGiven mass=0.0263g

Substitute molar mass of F , molarity and volume of solution in equation (3) as,

Molarity=Given massMolar massVolume of solution in litres0.070×103=Given mass19.0gmol1.0LGiven mass=0.0013g

The mass of all the ions is tabulated as,

Ions Mass of solute (g)
Na+ 11.048
K+ 0.4090
Mg2+ 1.316
Ca2+ 0.4220
Sr2+ 0.0081
Cl 19.832
SO42 2.779
HCO3 0.129
Br 0.0691
B(OH)3 0.0263
F 0.0013

Table 3

The conversion of g to kg is done as,

1g=0.001kg

Hence, the conversion of 1.022g to kg is done as,

1.022g=1.022×0.001kg=1.022×103kg

Similarly, the conversion of 11.048g to kg is done as,

11.048g=11.048×0.001kg=11.048×103kg

Similarly, the conversion of 0.4090g to kg is done as,

0.4090g=0.4090×0.001kg=0.4090×103kg

Similarly, the conversion of 1.316g to kg is done as,

1.316g=1.316×0.001kg=1.316×103kg

Similarly, the conversion of 0.4220g to kg is done as,

0.4220g=0.4220×0.001kg=0.4220×103kg

Similarly, the conversion of 0.0081g to kg is done as,

0.0081g=0.0081×0.001kg=0.0081×103kg

Similarly, the conversion of 19.832g to kg is done as,

19.832g=19.832×0.001kg=19.832×103kg

Similarly, the conversion of 2.779g to kg is done as,

2.779g=2.779×0.001kg=2.779×103kg

Similarly, the conversion of 0.129g to kg is done as,

0.129g=0.129×0.001kg=0.129×103kg

Similarly, the conversion of 0.0691g to kg is done as,

0.0691g=0.0691×0.001kg=0.0691×103kg

Similarly, the conversion of 0.0263g to kg is done as,

0.0263g=0.0263×0.001kg=0.0263×103kg

Similarly, the conversion of 0.0013g to kg is done as,

0.0013g=0.0013×0.001kg=0.0013×103kg

The conversion of mL to L is done as,

1mL=0.001L

The mass of 0.001L solution =1.022×103kg

The mass of 1L solution =1.022×1030.001kg=1.022kg

Mass of solvent is given as,

Mass of solvent=Mass of solutionMass of solute (4)

Substitute the mass of solution and mass of Na+ solute in equation (4),

Mass of solvent=Mass of solutionMass of solute=1022×103kg11.048×103kg=1.011kg

Substitute the mass of solution and mass of K+ solute in equation (4),

Mass of solvent=Mass of solutionMass of solute=1022×103kg0.4090×103kg=1.022kg

Substitute the mass of solution and mass of Mg2+ solute in equation (4),

Mass of solvent=Mass of solutionMass of solute=1022×103kg1.316×103kg=1.021kg

Substitute the mass of solution and mass of Ca2+ solute in equation (4),

Mass of solvent=Mass of solutionMass of solute=1022×103kg0.4220×103kg=1.022kg

Substitute the mass of solution and mass of Sr2+ solute in equation (4),

Mass of solvent=Mass of solutionMass of solute=1022×103kg0.0081×103kg=1.022kg

Substitute the mass of solution and mass of Cl solute in equation (4),

Mass of solvent=Mass of solutionMass of solute=1022×103kg19.832×103kg=1.002kg

Substitute the mass of solution and mass of SO42 solute in equation (4),

Mass of solvent=Mass of solutionMass of solute=1022×103kg2.779×103kg=1.02kg

Substitute the mass of solution and mass of HCO3 solute in equation (4),

Mass of solvent=Mass of solutionMass of solute=1022×103kg0.129×103kg=1.02kg

Substitute the mass of solution and mass of Br solute in equation (4),

Mass of solvent=Mass of solutionMass of solute=1022×103kg0.0691×103kg=1.02kg

Substitute the mass of solution and mass of B(OH)3 solute in equation (4),

Mass of solvent=Mass of solutionMass of solute=1022×103kg0.0263×103kg=1.02kg

Substitute the mass of solution and mass of F solute in equation (4),

Mass of solvent=Mass of solutionMass of solute=1022×103kg0.0013×103kg=1.02kg

The mass of solvent for the given ions that act as solute is tabulated as,

Ions Mass of solvent (kg)
Na+ 1.011
K+ 1.022
Mg2+ 1.021
Ca2+ 1.022
Sr2+ 1.022
Cl 1.002
SO42 1.02
HCO3 1.02
Br 1.02
B(OH)3 1.02
F 1.02

Table 4

The molality of the solution is given as,

Molality=Numbe of moles of soluteMass of solvent in kg (5)

Substitute the value of number of moles of Na+ and mass of solvent in equation (5).

Molality=480.57×103mol1.011kg=0.4753mol/kg_

Substitute the value of number of moles of K+ and mass of solvent in equation (5).

Molality=10.46×103mol1.022kg=1.02×10-2mol/kg_

Substitute the value of number of moles of Mg2+ and mass of solvent in equation (5).

Molality=54.14×103mol1.021kg=5.3×10-2mol/kg_

Substitute the value of number of moles of Ca2+ and mass of solvent in equation (5).

Molality=10.53×103mol1.022kg=1.03×10-2mol/kg_

Substitute the value of number of moles of Sr2+ and mass of solvent in equation (5).

Molality=0.0928×103mol1.022kg=9.08×10-5mol/kg_

Substitute the value of number of moles of Cl and mass of solvent in equation (5).

Molality=559.40×103mol1.022kg=5.5×10-1mol/kg_

Substitute the value of number of moles of SO42 and mass of solvent in equation (5).

Molality=28.93×103mol1.022kg=2.83×10-2mol/kg_

Substitute the value of number of moles of HCO3 and mass of solvent in equation (5).

Molality=2.11×103mol1.022kg=2.06×10-3mol/kg_

Substitute the value of number of moles of Br and mass of solvent in equation (5).

Molality=0.865×103mol1.022kg=8.5×10-4mol/kg_

Substitute the value of number of moles of B(OH)3 and mass of solvent in equation (5).

Molality=0.426×103mol1.022kg=4.2×10-4mol/kg_

Substitute the value of number of moles of F and mass of solvent in equation (5).

Molality=0.070×103mol1.022kg=6.9×10-5mol/kg_

The molality of the given ions is tabulated as,

Ions Molality (mol/kg)
Na+ 0.4753
K+ 1.02×10-2
Mg2+ 5.3×10-2
Ca2+ 1.03×10-2
Sr2+ 9.08×10-5
Cl 5.5×10-1
SO42 2.83×10-2
HCO3 2.06×10-3
Br 8.5×10-4
B(OH)3 4.2×10-4
F 6.9×10-5

Table 4

Conclusion

The molality for the given ions of sea water is tabulated as,

Ions Molality (mol/kg)
Na+ 0.4753
K+ 1.02×10-2
Mg2+ 5.3×10-2
Ca2+ 1.03×10-2
Sr2+ 9.08×10-5
Cl 5.5×10-1
SO42 2.83×10-2
HCO3 2.06×10-3
Br 8.5×10-4
B(OH)3 4.2×10-4
F 6.9×10-5

Table 4

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Chemistry

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