Chemistry: An Atoms-Focused Approach
Chemistry: An Atoms-Focused Approach
14th Edition
ISBN: 9780393912340
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster
Publisher: W. W. Norton & Company
Question
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Chapter 11, Problem 11.40QA
Interpretation Introduction

To calculate

The molality of each of the given seawater ions from given molarity and density.

Expert Solution & Answer
Check Mark

Answer to Problem 11.40QA

Solution

a) Molality of Na+ = 0.474 m

b) Molality of K+ = 1.02 x 10-2 m

c) Molality of Mg2+ = 5.29 x 10-2 m

d) Molality of Ca2+ = 1.03 x 10-2 m

e) Molality of Sr2+ = 9.05 x 10-5 m

f)  Molality of Cl- = 0.557 m

g) Molality of SO42- = 2.83 x 10-2 m

h) Molality of HCO3- = 2.06 x 10-3 m

i) Molality of Br - = 8.44 x 10-4 m

j) Molality of B(OH)3 = 4.16 x 10-4 m

k) Molality of F - = 6.83 x 10-5 m

Explanation of Solution

Molality of the solution is moles of solute divided by the mass of solvent in kg.

molality= mol of solutekg of solvent

Molarity of the solution is moles of solute divide by L of solution.

molarity= mol of soluteL of solution

Density of solution is mass divided by volume of solution

Density= mass of solutionvolume of solution

Mass of the solution is mass of solvent and mass of solute

Mass of solution=mass of solute+mass of solvent

To calculate the molality of the ions, we need to calculate the moles of solute and mass of solvent in kg from the given density and molarity.

a) Calculation of molality of Na+

Given

Molarity = 480.57 mM Na+= 480.57 mM Na+ × 1 M1000 mM=0.48057 M Na+

Density (d) = 1.025 g/mL

Calculations

To calculate the moles of Na+ from 0.48057 M, we need to assume that the volume of solution is 1 L.

0.48057 M Na+= mol of Na+1 L of solution

1 L solution ×  0.48057 mol of Na+1 L of solution =0.48057 mol Na+

So moles of solute = 0.48057 mol Na+

The mass of Na+ is

0.48057 mol Na+ × 22.99 g Na+1 mol Na+ =11.048 g Na+

The density of the solution in g/mL is

1.025 gmL× 1000 mL1 L=1025 gL

So 1 L of the solution contains 1025 g of solution.

We know the mass of Na+ and mass of solution; we can calculate the mass of solvent.

Mass of solution=mass of solute+mass of solvent

mass of solvent=1025 g-11.048 g=1013.952g

1013.952 g× 1 kg1000 g=1.013952 kg

The molality of the Na+ ion in sea water from the moles and mass of solvent is

molality= 0.48057 mol of Na+1.013952 kg of solvent =0.474 m

The molality of the Na+ ion in sea water is 0.474 m.

b) Calculation of molality of K+

Given

Molarity = 10.46 mM K+=  10.46 mM K+ × 1 M1000 mM=0.01046 M K+

Density (d) = 1.025 g/mL

Calculations

To calculate the moles of K+ from 0.01046 M, we need to assume that the volume of solution is 1 L.

0.01046 M K+= mol of K+1 L of solution

1 L solution ×  0.01046 mol of K+1 L of solution =0.01046 mol K+

So moles of solute = 0.01046 mol K+

The mass of K+ is

0.01046 mol K+ × 39.10 g K+1 mol K+ =0.4090 g K+

The density of the solution in g/mL is

1.025 gmL× 1000 mL1 L=1025 gL

So 1 L of the solution contains 1025 g of solution.

We know the mass of K+ and mass of solution; we can calculate the mass of solvent.

Mass of solution=mass of solute+mass of solvent

mass of solvent=1025 g-0.4090 g=1024.591 g solvent

1024.591 g water× 1 kg1000 g=1.0246 kg water

The molality of the K+ ions from the moles and mass of solvent is

molality= 0.01046 mol of K+1.0246 kg of solvent =0.01021 m=1.02 × 10-2 m

The molality of the K+ ion is 1.02 × 10-2 m.

c) Calculation of molality of Mg2+

Given

Molarity = 54.14 mM Mg2+=54.14 mM Mg2+ × 1 M1000 mM=0.05414 M Mg2+

Density (d) = 1.025 g/mL

Calculations

To calculate the moles of Mg2+ from 0.05414 M, we need to assume that the volume of solution is 1 L.

0.05414 M Mg2+= moles of Mg2+1 L of solution

1 L solution ×  0.05414 mol of Mg2+1 L of solution =0.05414 mol Mg2+

So moles of solute = 0.05414 mol Mg2+

The mass of Mg2+ is

0.05414 mol Mg2+ × 24.31 g Mg2+1 mol Mg2+ =1.3161 g Mg2+

The density of the solution in g/mL is

1.025 gmL× 1000 mL1 L=1025 gL

So 1 L of the solution contains 1025 g of solution.

We know the mass of Mg2+ and mass of solution; we can calculate the mass of solvent.

Mass of solution=mass of solute+mass of solvent

mass of solvent=1025 g-1.3161 g=1023.684 g solvent

1023.684 g solvent× 1 kg1000 g=1.0237 kg solvent

The molality of the Mg2+ ions from the moles and mass of solvent is

molality= 0.05414 mol of Mg2+1.0237 kg of solvent =0.05289 m=5.29 × 10-2 m

The molality of the Mg2+ ion is 5.29 × 10-2 m.

d) Calculation of molality of Ca2+

Given

Molarity = 10.53 mM Ca2+= 10.53 mM Ca2+ × 1 M1000 mM=0.01053 M Ca2+

Density (d) = 1.025 g/mL

Calculations

To calculate the moles of Ca2+ from 0.01053 M, we need to assume that the volume of solution is 1 L.

0.01053M Ca2+= mol of Ca2+1 L of solution

1 L solution ×  0.01053 mol of Ca2+1 L of solution =0.01053 mol Ca2+

So moles of solute = 0.01053 mol Ca2+

The mass of Ca2+ is

0.01053 mol Ca2+ × 40.08 g Ca2+1 mol Ca2+ =0.4220 g Ca2+

The density of the solution in g/mL is

1.025 gmL× 1000 mL1 L=1025 gL

So 1 L of the solution contains 1025 g of solution.

We know the mass of Ca2+ and mass of solution; we can calculate the mass of solvent.

Mass of solution=mass of solute+mass of solvent

mass of solvent=1025 g-0.4220 g=1024.578 g solvent

1024.578 g solvent× 1 kg1000 g=1.0246 kg solvent

The molality of the Ca2+ ions from the moles and mass of solvent is

molality= 0.01053 mol of Ca2+1.0246 kg of solvent =0.01028 m=1.03 × 10-2 m

The molality of the Ca2+ ion is 1.03 × 10-2 m.

e) Calculation of molality of Sr2+

Given

Molarity = 0.0928 mM Sr2+= 0.0928 mM Sr2+ × 1 M1000 mM=0.0000928 M Sr2+

Density (d) = 1.025 g/mL

Calculations

To calculate the moles of Sr2+ from 0.0000928 M, we need to assume that the volume of solution is 1 L.

0.0000928 M Sr2+= mol of Sr2+1 L of solution

1 L solution ×  0.0000928 mol of Sr2+1 L of solution =0.0000928 mol Sr2+

So moles of solute = 0.0000928 mol Sr2+

The mass of Sr2+ is

0.0000928 mol Sr2+ × 87.62 g Sr2+1 mol Sr2+ =0.008131 g Sr2+

The density of the solution in g/mL is

1.025 gmL× 1000 mL1 L=1025 gL

So 1 L of the solution contains 1025 g of solution.

We know the mass of Sr2+ and mass of solution; we can calculate the mass of solvent.

Mass of solution=mass of solute+mass of solvent

mass of solvent=1025 g-0.008131 g=1024.992 g solvent

1024.992 g solvent× 1 kg1000 g=1.024992 kg solvent

The molality of the Sr2+ ions from the moles and mass of solvent is

molality= 0.0000928 mol of Sr2+1.024992 kg of solvent =0.00009054 m=9.05 × 10-5 m

The molality of the Sr2+ ion is 9.05 × 10-5 m.

f) Calculation of molality of Cl-

Given

Molarity = 559.40 mM Cl-= 559.40 mM Cl- × 1 M1000 mM=0.5594 M Cl-

Density (d) = 1.025 g/mL

Calculations

To calculate the moles of Cl- from 0.5594 M, we need to assume that the volume of solution is 1 L.

0.5594 M Cl-= mol of  Cl-1 L of solution

1 L solution ×  0.5594 mol of  Cl-1 L of solution =0.5594 mol  Cl-

So moles of solute = 0.5594 mol Cl-

The mass of Cl- is

0.5594 mol  Cl- × 35.45 g  Cl-1 mol  Cl- =19.8307 g  Cl-

The density of the solution in g/mL is

1.025 gmL× 1000 mL1 L=1025 gL

So 1 L of the solution contains 1025 g of solution.

We know the mass of Cl- and mass of solution; we can calculate the mass of solvent.

Mass of solution=mass of solute+mass of solvent

mass of solvent=1025 g-19.8307 g=1005.169 g solvent

1005.169 g solvent× 1 kg1000 g=1.0052 kg solvent

The molality of the Cl- ions from the moles and mass of solvent is

molality= 0.5594 mol of  Cl-1.0052 kg of solvent =0.557 m=5.57 × 10-1 m

The molality of the Cl- ion is 5.57 × 10-1 m.

g) Calculation of molality of SO42-

Given

Molarity = 28.93 mM SO42-= 28.93 mM SO42- × 1 M1000 mM=0.02893 M SO42-

Density (d) = 1.025 g/mL

Calculations

To calculate the moles of SO42- from 0.02893 M, we need to assume that the volume of solution is 1 L.

0.02893 M SO42-= mol of  SO42-1 L of solution

1 L solution ×  0.02893 mol of  SO42-1 L of solution =0.02893 mol  SO42-

So moles of solute = 0.02893 mol SO42-

The mass of SO42- is

0.02893 mol  SO42- × 96.06 g  SO42-1 mol  SO42- =2.7790 g  SO42-

The density of the solution in g/mL is

1.025 gmL× 1000 mL1 L=1025 gL

So 1 L of the solution contains 1025 g of solution.

We know the mass of SO42- and mass of solution; we can calculate the mass of solvent.

Mass of solution=mass of solute+mass of solvent

mass of solvent=1025 g-2.7790 g=1022.221 g solvent

1022.221 g solvent× 1 kg1000 g=1.02222 kg solvent

The molality of the SO42- ions from the moles and mass of solvent is

molality= 0.02893 mol of  Cl-1.02222 kg of solvent =0.02830 m=2.83 × 10-2 m

The molality of the SO42- ion is 2.83 × 10-2 m.

h) Calculation of molality of HCO3-

Given

Molarity = 2.11 mMHCO3-= 2.11 mMHCO3- × 1 M1000 mM=0.00211 M HCO3-

Density (d) = 1.025 g/mL

Calculations

To calculate the moles of HCO3- from 0.00211 M, we need to assume that the volume of solution is 1 L.

0.00211 MHCO3-= mol of  HCO3-1 L of solution

1 L solution ×  0.00211 mol of  HCO3-1 L of solution =0.00211mol  HCO3-

So moles of solute = 0.00211 mol HCO3-

The mass of HCO3-  is

0.00211 mol  HCO3- × 61.02 g HCO3-1 mol  HCO3- =0.128752 g  HCO3-

The density of the solution in g/mL is

1.025 gmL× 1000 mL1 L=1025 gL

So 1 L of the solution contains 1025 g  of solution.

We know the mass of HCO3- and mass of solution; we can calculate the mass of solvent.

Mass of solution=mass of solute+mass of solvent

mass of solvent=1025 g-0.128752 g=1024.871 g solvent

1024.871g solvent× 1 kg1000 g=1.0249 kg solvent

The molality of the HCO3- ions from the moles and mass of solvent is

molality= 0.00211 mol of HCO3-1.0249 kg of solvent =0.002059 m=2.06 × 10-3 m

The molality of the HCO3- ion is 2.06 × 10-3 m.

i) Calculation of molality of Br-

Given

Molarity = 0.865 mM Br-= 0.865 mM Br-  × 1 M1000 mM=0.000865 M Br-

Density (d) = 1.025 g/mL

Calculations

To calculate the moles of Br- from 0.000865 M, we need to assume that the volume of solution is 1 L.

0.000865 MBr-= mol of  Br-1 L of solution

1 L solution ×  0.000865 mol of  Br-1 L of solution =0.0008651 mol Br-

So moles of solute = 0.000865 mol Br-

The mass of Br-  is

0.000865 mol  Br- × 79.90 g Br-1 mol  Br- =0.069114 g  Br-

The density of the solution in g/mL is

1.025 gmL× 1000 mL1 L=1025 gL

So 1 L of the solution contains 1025 g of solution.

We know the mass of Br- and mass of solution; we can calculate the mass of solvent.

Mass of solution=mass of solute+mass of solvent

mass of solvent=1025 g-0.069114 g=1024.931 g solvent

1024.931 g solvent× 1 kg1000 g=1.0249 kg solvent

The molality of the Br- ions from the moles and mass of solvent is

molality= 0.000865 mol of Br-1.0249 kg of solvent =0.000844 m=8.44 × 10-4 m

The molality of the Br- ion is 8.44 × 10-4 m.

j) Calculation of molality of B(OH)3

Given

Molarity = 0.426 mM BOH3= 0.426 mM BOH3  × 1 M1000 mM=0.000426 mol BOH3

Density (d) = 1.025 g/mL

Calculations

To calculate the moles of BOH3 from 0.000426 M, we need to assume that the volume of solution is 1 L.

0.000426 M BOH3= mol of  BOH31 L of solution

1 L solution ×  0.000426 mol of  BOH31 L of solution =0.000426 mol BOH3

So moles of solute = 0.000426 mol BOH3

The mass of BOH3 is

0.000426 mol  BOH3 × 61.83 g BOH31 mol  Br- =0.02634 g  BOH3

The density of the solution in g/mL is

1.025 gmL× 1000 mL1 L=1025 gL

So 1 L of the solution contains 1025 g of solution.

We know the mass of BOH3 and mass of solution; we can calculate the mass of solvent.

Mass of solution=mass of solute+mass of solvent

mass of solvent=1025 g-0.02634 g=1024.974 g solvent

1024.974 g solvent× 1 kg1000 g=1.0249 kg solvent

The molality of the BOH3 from the moles and mass of solvent is

molality= 0.000426 mol of BOH31.0249 kg of solvent =0.000416 m=4.16 × 10-4 m

The molality of the BOH3 is 4.16 × 10-4 m.

i) Calculation of molality of F-

Given

Molarity = 0.070 mM F-= 0.070 mM F-  × 1 M1000 mM=0.000070 M F-

Density (d) = 1.025 g/mL

Calculations

To calculate the moles of F- from 0.000070 M, we need to assume that the volume of solution is 1 L.

0.000070 MF-= mol of  F-1 L of solution

1 L solution ×  0.000070 mol of  F-1 L of solution =0.000070 mol F-

So moles of solute = 0.000070 mo l F-

The mass of F-  is

0.000070 mol  F- × 19.00 g F-1 mol  F- =0.00133 g  F-

The density of the solution in g/mL is

1.025 gmL× 1000 mL1 L=1025 gL

So 1 L of the solution contains 1025 g of solution.

We know the mass of Br- and mass of solution; we can calculate the mass of solvent.

Mass of solution=mass of solute+mass of solvent

mass of solvent=1025 g-0.00133 g=1024.999 g solvent

1024.999g solvent× 1 kg1000 g=1.024999 kg solvent

The molality of the F- ions from the moles and mass of solvent is

molality= 0.000070 mol of F-1.024999 kg of solvent =0.0000683 m=6.83 × 10-5 m

The molality of the F- ion is 6.83 × 10-5 m.

Conclusion

Molarity of the major ions present sea water has been calculated from the molarity and density.

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Chapter 11 Solutions

Chemistry: An Atoms-Focused Approach

Ch. 11 - Prob. 11.12QACh. 11 - Prob. 11.13QACh. 11 - Prob. 11.14QACh. 11 - Prob. 11.15QACh. 11 - Prob. 11.16QACh. 11 - Prob. 11.17QACh. 11 - Prob. 11.18QACh. 11 - Prob. 11.19QACh. 11 - Prob. 11.20QACh. 11 - Prob. 11.21QACh. 11 - Prob. 11.22QACh. 11 - Prob. 11.23QACh. 11 - Prob. 11.24QACh. 11 - Prob. 11.25QACh. 11 - Prob. 11.26QACh. 11 - Prob. 11.27QACh. 11 - Prob. 11.28QACh. 11 - Prob. 11.29QACh. 11 - Prob. 11.30QACh. 11 - Prob. 11.31QACh. 11 - Prob. 11.32QACh. 11 - Prob. 11.33QACh. 11 - Prob. 11.34QACh. 11 - Prob. 11.35QACh. 11 - Prob. 11.36QACh. 11 - Prob. 11.37QACh. 11 - Prob. 11.38QACh. 11 - Prob. 11.39QACh. 11 - Prob. 11.40QACh. 11 - Prob. 11.41QACh. 11 - Prob. 11.42QACh. 11 - Prob. 11.43QACh. 11 - Prob. 11.44QACh. 11 - Prob. 11.45QACh. 11 - Prob. 11.46QACh. 11 - Prob. 11.47QACh. 11 - Prob. 11.48QACh. 11 - Prob. 11.49QACh. 11 - Prob. 11.50QACh. 11 - Prob. 11.51QACh. 11 - Prob. 11.52QACh. 11 - Prob. 11.53QACh. 11 - Prob. 11.54QACh. 11 - Prob. 11.55QACh. 11 - Prob. 11.56QACh. 11 - Prob. 11.57QACh. 11 - Prob. 11.58QACh. 11 - Prob. 11.59QACh. 11 - Prob. 11.60QACh. 11 - Prob. 11.61QACh. 11 - Prob. 11.62QACh. 11 - Prob. 11.63QACh. 11 - Prob. 11.64QACh. 11 - Prob. 11.65QACh. 11 - Prob. 11.66QACh. 11 - Prob. 11.67QACh. 11 - Prob. 11.68QACh. 11 - Prob. 11.69QACh. 11 - Prob. 11.70QACh. 11 - Prob. 11.71QACh. 11 - Prob. 11.72QACh. 11 - Prob. 11.73QACh. 11 - Prob. 11.74QACh. 11 - Prob. 11.75QACh. 11 - Prob. 11.76QACh. 11 - Prob. 11.77QACh. 11 - Prob. 11.78QACh. 11 - Prob. 11.79QACh. 11 - Prob. 11.80QACh. 11 - Prob. 11.81QACh. 11 - Prob. 11.82QACh. 11 - Prob. 11.83QACh. 11 - Prob. 11.84QA
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