The amount of reagents required and range of osmotic pressure has to be calculated. Concept Introduction: The mass of the compound is calculated by taking the products of molar mass of the compound to the given mass. The mass of compound can be given by, Mass of compound(in grams)=Molar mass (in g)×Given mass(in g)

Question

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Answer 1

Question

Chapter 11, Problem 108AE

Interpretation Introduction

Interpretation:

The amount of reagents required and range of osmotic pressure has to be calculated.

Concept Introduction:

The mass of the compound is calculated by taking the products of molar mass of the compound to the given mass. The mass of compound can be given by,

$Mass of compound(in grams)=Molar mass (in g)×Given mass(in g)$

Expert Solution

Answer to Problem 108AE

$305.9 mg$ Sodium lactate,

$20.0 mg CaCl2.2H2O$ ,

$29.9 mg KCl$ and

$601.2 mg NaCl$ are required to prepare lactated Ringer’s reagent.

To calculate the mass of individual elements

Explanation of Solution

Mass of Sodium = $285-315 mg$

Mass of Potassium = $14.1-17.3 mg$

Mass of Calcium = $4.9-6.0 mg$

Mass of Chlorine = $368-408 mg$

Mass of Lactate = $231-261 mg$

To calculate the mass of individual elements

Molar mass of Sodium lactate = $112.06 mg$

Molar mass of Lactate = $89.07 mg$

Molar mass of $CaCl2.2H2O$ = $147.01 mg$

Molecular mass of Calcium = $40.08 mg$

Molar mass of $KCl$ = $74.55 mg$

Molecular weight of Potassium = $39.10 mg$

Molar mass of $NaCl$ = $58.44 mg$

Molecular mass of Sodium= $22.99 mg$

The average values for each ion are,

$300.00 mg Na+,15.7 mg K+,5.45 mg Ca2+,388 mg Cl- and 246 mg Lactate$

The source of Lactate is $NaC3H5O3$

Mass of Lactate = $246 mg C3H5O3-×112.06 mg NaC3H5O389.07 mg C3H5O3-=309.5 mg$

The source of $Ca2+$ is $CaCl2.2H2O$

Mass of $CaCl2.2H2O$ = $5.45 mg Ca2+×147.01 mg CaCl2.2H2O40.08 mg Ca2+=20 mg CaCl2.2H2O$

The source of $K+$ is $KCl$

Mass of $KCl$ = $15.7 mg K+×74.55 mg KCl39.10 mg K+=29.9 mg$

Mass of $Na+$ added = $309.5 mg Sodium lactate-246.0 mg Lactate=63.5 mg Na+$

Additional amount of Sodium $236.5 mg Na+$ is added to get desired $300 mg$

Mass of Sodium added = $236.5 mg Na+×58.44 mg NaCl22.99 mg Na+=601.2 mg$

Mass of $Cl-$ added = $20.0 mg CaCl2.2H2O×70.90 mg Cl-147.01 mg CaCl2.2H2O=9.6 mg$

$20.0 mg CaCl2.2H2O =9.6 mg Cl-29.9 mg KCl-15.7 mg K+ =14.2 mg Cl-601.2 mg NaCl-236.5 mg Na+=364.7 mg Cl-$

Total $Cl-$ = $388.5 mg$

Therefore,

$305.9 mg$ Sodium lactate, $20.0 mg CaCl2.2H2O$ , $29.9 mg KCl$ and $601.2 mg NaCl$ are required to prepare lactated Ringer’s reagent.

Conclusion

The mass of individual elements was calculated using their respective molar mass and molecular weight and the given weight. A typical analytical balance can nearly measure to $0.1 mg$ . Therefore, $305.9 mg$ Sodium lactate, $20.0 mg CaCl2.2H2O$ , $29.9 mg KCl$ and $601.2 mg NaCl$ are required to prepare lactated Ringer’s reagent.

Interpretation Introduction

Interpretation:

The osmotic pressure at minimum and maximum concentration has to be calculated.

Concept Introduction:

Colligative properties of a substance include the depression in the freezing point, elevation of boiling-point and osmotic pressure. These are dependant only on the number present and not based on the solute particles present in an ideal solution.

The osmotic pressure can be given by the equation,

$Π=MRT$

$Here,Π=Osmotic pressure M=Molarity of solution R= Gas law constant T=Temperature$

Expert Solution

Answer to Problem 108AE

The range of osmotic pressure is $6.59 atm to 7.30 atm$

Explanation of Solution

To calculate the osmotic pressure at minimum and maximum concentration

Record the given info

Mass of Sodium = $285-315 mg$

Mass of Potassium = $14.1-17.3 mg$

Mass of Calcium = $4.9-6.0 mg$

Mass of Chlorine = $368-408 mg$

Mass of Lactate = $231-261 mg$

The masses of individual elements present in the reagent are recorded as shown above.

To calculate the minimum and maximum concentrations of ions

Molar mass of Lactate = $89.07 mg$

Molecular mass of Calcium = $40.08 mg$

Molecular weight of Potassium = $39.10 mg$

Molecular mass of Sodium= $22.99 mg$

At minimum concentration,

Molarity of Sodium = $285 mg Na+100. mL×1 mmol22.99 mg=0.124 M$

Molarity of Potassium = $14.1 mg K+100.mL×1 mmol39.10 mg=0.00361 M$

Molarity of Lactate = $231 mg C3H5O3-100.mL×1 mmol89.07 mg=0.0259 M$

Molarity of Calcium = $4.9 mg Ca2+100.mL×1 mmol40.08 mg=0.0012 M$

Molarity of Chlorine = $368 mg Cl-100.mL×1 mmol35.45 mg=0.104 M$

The total concentration = $0.124+0.00361+0.0012+0.104+0.0259$

= $0.259 M$

At maximum concentration,

Molarity of Sodium = $315 mg Na+100. mL×1 mmol22.99 mg=0.137 M$

Molarity of Potassium = $17.3 mg K+100.mL×1 mmol39.10 mg=0.00442 M$

Molarity of Lactate = $261 mg C3H5O3-100.mL×1 mmol89.07 mg=0.0293 M$

Molarity of Calcium = $6.0 mg Ca2+100.mL×1 mmol40.08 mg=0.0015 M$

Molarity of Chlorine = $408 mg Cl-100.mL×1 mmol35.45 mg=0.115 M$

The total concentration= $0.137+0.00442+0.0015+0.115+0.0293$

= $0.287 M$

To calculate the osmotic pressure at minimum and maximum concentration

At minimum concentration,

$π=MRT =0.259 molL×0.08206 L atmK mol×310.K =6.59 atm$

At maximum concentration,

$π=MRT =0.287 molL×0.08206 L atmK mol×310.K =7.30 atm$

At minimum concentration, osmotic pressure= $6.59 atm$

At maximum concentration, osmotic pressure= $7.30 atm$

Conclusion

The osmotic pressure at minimum and maximum concentrations was calculated using the molarities at minimum and maximum concentration. The osmotic pressure at minimum and maximum concentrations were $6.59 atm$ and $7.30 atm$ .

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Answer 2

Question

Chapter 11, Problem 108AE

Interpretation Introduction

Interpretation:

The amount of reagents required and range of osmotic pressure has to be calculated.

Concept Introduction:

The mass of the compound is calculated by taking the products of molar mass of the compound to the given mass. The mass of compound can be given by,

$Mass of compound(in grams)=Molar mass (in g)×Given mass(in g)$

Expert Solution

Answer to Problem 108AE

$305.9 mg$ Sodium lactate,

$20.0 mg CaCl2.2H2O$ ,

$29.9 mg KCl$ and

$601.2 mg NaCl$ are required to prepare lactated Ringer’s reagent.

To calculate the mass of individual elements

Explanation of Solution

Mass of Sodium = $285-315 mg$

Mass of Potassium = $14.1-17.3 mg$

Mass of Calcium = $4.9-6.0 mg$

Mass of Chlorine = $368-408 mg$

Mass of Lactate = $231-261 mg$

To calculate the mass of individual elements

Molar mass of Sodium lactate = $112.06 mg$

Molar mass of Lactate = $89.07 mg$

Molar mass of $CaCl2.2H2O$ = $147.01 mg$

Molecular mass of Calcium = $40.08 mg$

Molar mass of $KCl$ = $74.55 mg$

Molecular weight of Potassium = $39.10 mg$

Molar mass of $NaCl$ = $58.44 mg$

Molecular mass of Sodium= $22.99 mg$

The average values for each ion are,

$300.00 mg Na+,15.7 mg K+,5.45 mg Ca2+,388 mg Cl- and 246 mg Lactate$

The source of Lactate is $NaC3H5O3$

Mass of Lactate = $246 mg C3H5O3-×112.06 mg NaC3H5O389.07 mg C3H5O3-=309.5 mg$

The source of $Ca2+$ is $CaCl2.2H2O$

Mass of $CaCl2.2H2O$ = $5.45 mg Ca2+×147.01 mg CaCl2.2H2O40.08 mg Ca2+=20 mg CaCl2.2H2O$

The source of $K+$ is $KCl$

Mass of $KCl$ = $15.7 mg K+×74.55 mg KCl39.10 mg K+=29.9 mg$

Mass of $Na+$ added = $309.5 mg Sodium lactate-246.0 mg Lactate=63.5 mg Na+$

Additional amount of Sodium $236.5 mg Na+$ is added to get desired $300 mg$

Mass of Sodium added = $236.5 mg Na+×58.44 mg NaCl22.99 mg Na+=601.2 mg$

Mass of $Cl-$ added = $20.0 mg CaCl2.2H2O×70.90 mg Cl-147.01 mg CaCl2.2H2O=9.6 mg$

$20.0 mg CaCl2.2H2O =9.6 mg Cl-29.9 mg KCl-15.7 mg K+ =14.2 mg Cl-601.2 mg NaCl-236.5 mg Na+=364.7 mg Cl-$

Total $Cl-$ = $388.5 mg$

Therefore,

$305.9 mg$ Sodium lactate, $20.0 mg CaCl2.2H2O$ , $29.9 mg KCl$ and $601.2 mg NaCl$ are required to prepare lactated Ringer’s reagent.

Conclusion

The mass of individual elements was calculated using their respective molar mass and molecular weight and the given weight. A typical analytical balance can nearly measure to $0.1 mg$ . Therefore, $305.9 mg$ Sodium lactate, $20.0 mg CaCl2.2H2O$ , $29.9 mg KCl$ and $601.2 mg NaCl$ are required to prepare lactated Ringer’s reagent.

Interpretation Introduction

Interpretation:

The osmotic pressure at minimum and maximum concentration has to be calculated.

Concept Introduction:

Colligative properties of a substance include the depression in the freezing point, elevation of boiling-point and osmotic pressure. These are dependant only on the number present and not based on the solute particles present in an ideal solution.

The osmotic pressure can be given by the equation,

$Π=MRT$

$Here,Π=Osmotic pressure M=Molarity of solution R= Gas law constant T=Temperature$

Expert Solution

Answer to Problem 108AE

The range of osmotic pressure is $6.59 atm to 7.30 atm$

Explanation of Solution

To calculate the osmotic pressure at minimum and maximum concentration

Record the given info

Mass of Sodium = $285-315 mg$

Mass of Potassium = $14.1-17.3 mg$

Mass of Calcium = $4.9-6.0 mg$

Mass of Chlorine = $368-408 mg$

Mass of Lactate = $231-261 mg$

The masses of individual elements present in the reagent are recorded as shown above.

To calculate the minimum and maximum concentrations of ions

Molar mass of Lactate = $89.07 mg$

Molecular mass of Calcium = $40.08 mg$

Molecular weight of Potassium = $39.10 mg$

Molecular mass of Sodium= $22.99 mg$

At minimum concentration,

Molarity of Sodium = $285 mg Na+100. mL×1 mmol22.99 mg=0.124 M$

Molarity of Potassium = $14.1 mg K+100.mL×1 mmol39.10 mg=0.00361 M$

Molarity of Lactate = $231 mg C3H5O3-100.mL×1 mmol89.07 mg=0.0259 M$

Molarity of Calcium = $4.9 mg Ca2+100.mL×1 mmol40.08 mg=0.0012 M$

Molarity of Chlorine = $368 mg Cl-100.mL×1 mmol35.45 mg=0.104 M$

The total concentration = $0.124+0.00361+0.0012+0.104+0.0259$

= $0.259 M$

At maximum concentration,

Molarity of Sodium = $315 mg Na+100. mL×1 mmol22.99 mg=0.137 M$

Molarity of Potassium = $17.3 mg K+100.mL×1 mmol39.10 mg=0.00442 M$

Molarity of Lactate = $261 mg C3H5O3-100.mL×1 mmol89.07 mg=0.0293 M$

Molarity of Calcium = $6.0 mg Ca2+100.mL×1 mmol40.08 mg=0.0015 M$

Molarity of Chlorine = $408 mg Cl-100.mL×1 mmol35.45 mg=0.115 M$

The total concentration= $0.137+0.00442+0.0015+0.115+0.0293$

= $0.287 M$

To calculate the osmotic pressure at minimum and maximum concentration

At minimum concentration,

$π=MRT =0.259 molL×0.08206 L atmK mol×310.K =6.59 atm$

At maximum concentration,

$π=MRT =0.287 molL×0.08206 L atmK mol×310.K =7.30 atm$

At minimum concentration, osmotic pressure= $6.59 atm$

At maximum concentration, osmotic pressure= $7.30 atm$

Conclusion

The osmotic pressure at minimum and maximum concentrations was calculated using the molarities at minimum and maximum concentration. The osmotic pressure at minimum and maximum concentrations were $6.59 atm$ and $7.30 atm$ .

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Answer 3

Question

Chapter 11, Problem 108AE

Interpretation Introduction

Interpretation:

The amount of reagents required and range of osmotic pressure has to be calculated.

Concept Introduction:

The mass of the compound is calculated by taking the products of molar mass of the compound to the given mass. The mass of compound can be given by,

$Mass of compound(in grams)=Molar mass (in g)×Given mass(in g)$

Expert Solution

Answer to Problem 108AE

$305.9 mg$ Sodium lactate,

$20.0 mg CaCl2.2H2O$ ,

$29.9 mg KCl$ and

$601.2 mg NaCl$ are required to prepare lactated Ringer’s reagent.

To calculate the mass of individual elements

Explanation of Solution

Mass of Sodium = $285-315 mg$

Mass of Potassium = $14.1-17.3 mg$

Mass of Calcium = $4.9-6.0 mg$

Mass of Chlorine = $368-408 mg$

Mass of Lactate = $231-261 mg$

To calculate the mass of individual elements

Molar mass of Sodium lactate = $112.06 mg$

Molar mass of Lactate = $89.07 mg$

Molar mass of $CaCl2.2H2O$ = $147.01 mg$

Molecular mass of Calcium = $40.08 mg$

Molar mass of $KCl$ = $74.55 mg$

Molecular weight of Potassium = $39.10 mg$

Molar mass of $NaCl$ = $58.44 mg$

Molecular mass of Sodium= $22.99 mg$

The average values for each ion are,

$300.00 mg Na+,15.7 mg K+,5.45 mg Ca2+,388 mg Cl- and 246 mg Lactate$

The source of Lactate is $NaC3H5O3$

Mass of Lactate = $246 mg C3H5O3-×112.06 mg NaC3H5O389.07 mg C3H5O3-=309.5 mg$

The source of $Ca2+$ is $CaCl2.2H2O$

Mass of $CaCl2.2H2O$ = $5.45 mg Ca2+×147.01 mg CaCl2.2H2O40.08 mg Ca2+=20 mg CaCl2.2H2O$

The source of $K+$ is $KCl$

Mass of $KCl$ = $15.7 mg K+×74.55 mg KCl39.10 mg K+=29.9 mg$

Mass of $Na+$ added = $309.5 mg Sodium lactate-246.0 mg Lactate=63.5 mg Na+$

Additional amount of Sodium $236.5 mg Na+$ is added to get desired $300 mg$

Mass of Sodium added = $236.5 mg Na+×58.44 mg NaCl22.99 mg Na+=601.2 mg$

Mass of $Cl-$ added = $20.0 mg CaCl2.2H2O×70.90 mg Cl-147.01 mg CaCl2.2H2O=9.6 mg$

$20.0 mg CaCl2.2H2O =9.6 mg Cl-29.9 mg KCl-15.7 mg K+ =14.2 mg Cl-601.2 mg NaCl-236.5 mg Na+=364.7 mg Cl-$

Total $Cl-$ = $388.5 mg$

Therefore,

$305.9 mg$ Sodium lactate, $20.0 mg CaCl2.2H2O$ , $29.9 mg KCl$ and $601.2 mg NaCl$ are required to prepare lactated Ringer’s reagent.

Conclusion

The mass of individual elements was calculated using their respective molar mass and molecular weight and the given weight. A typical analytical balance can nearly measure to $0.1 mg$ . Therefore, $305.9 mg$ Sodium lactate, $20.0 mg CaCl2.2H2O$ , $29.9 mg KCl$ and $601.2 mg NaCl$ are required to prepare lactated Ringer’s reagent.

Interpretation Introduction

Interpretation:

The osmotic pressure at minimum and maximum concentration has to be calculated.

Concept Introduction:

Colligative properties of a substance include the depression in the freezing point, elevation of boiling-point and osmotic pressure. These are dependant only on the number present and not based on the solute particles present in an ideal solution.

The osmotic pressure can be given by the equation,

$Π=MRT$

$Here,Π=Osmotic pressure M=Molarity of solution R= Gas law constant T=Temperature$

Expert Solution

Answer to Problem 108AE

The range of osmotic pressure is $6.59 atm to 7.30 atm$

Explanation of Solution

To calculate the osmotic pressure at minimum and maximum concentration

Record the given info

Mass of Sodium = $285-315 mg$

Mass of Potassium = $14.1-17.3 mg$

Mass of Calcium = $4.9-6.0 mg$

Mass of Chlorine = $368-408 mg$

Mass of Lactate = $231-261 mg$

The masses of individual elements present in the reagent are recorded as shown above.

To calculate the minimum and maximum concentrations of ions

Molar mass of Lactate = $89.07 mg$

Molecular mass of Calcium = $40.08 mg$

Molecular weight of Potassium = $39.10 mg$

Molecular mass of Sodium= $22.99 mg$

At minimum concentration,

Molarity of Sodium = $285 mg Na+100. mL×1 mmol22.99 mg=0.124 M$

Molarity of Potassium = $14.1 mg K+100.mL×1 mmol39.10 mg=0.00361 M$

Molarity of Lactate = $231 mg C3H5O3-100.mL×1 mmol89.07 mg=0.0259 M$

Molarity of Calcium = $4.9 mg Ca2+100.mL×1 mmol40.08 mg=0.0012 M$

Molarity of Chlorine = $368 mg Cl-100.mL×1 mmol35.45 mg=0.104 M$

The total concentration = $0.124+0.00361+0.0012+0.104+0.0259$

= $0.259 M$

At maximum concentration,

Molarity of Sodium = $315 mg Na+100. mL×1 mmol22.99 mg=0.137 M$

Molarity of Potassium = $17.3 mg K+100.mL×1 mmol39.10 mg=0.00442 M$

Molarity of Lactate = $261 mg C3H5O3-100.mL×1 mmol89.07 mg=0.0293 M$

Molarity of Calcium = $6.0 mg Ca2+100.mL×1 mmol40.08 mg=0.0015 M$

Molarity of Chlorine = $408 mg Cl-100.mL×1 mmol35.45 mg=0.115 M$

The total concentration= $0.137+0.00442+0.0015+0.115+0.0293$

= $0.287 M$

To calculate the osmotic pressure at minimum and maximum concentration

At minimum concentration,

$π=MRT =0.259 molL×0.08206 L atmK mol×310.K =6.59 atm$

At maximum concentration,

$π=MRT =0.287 molL×0.08206 L atmK mol×310.K =7.30 atm$

At minimum concentration, osmotic pressure= $6.59 atm$

At maximum concentration, osmotic pressure= $7.30 atm$

Conclusion

The osmotic pressure at minimum and maximum concentrations was calculated using the molarities at minimum and maximum concentration. The osmotic pressure at minimum and maximum concentrations were $6.59 atm$ and $7.30 atm$ .

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Answer 4

Question

Chapter 11, Problem 108AE

Interpretation Introduction

Interpretation:

The amount of reagents required and range of osmotic pressure has to be calculated.

Concept Introduction:

The mass of the compound is calculated by taking the products of molar mass of the compound to the given mass. The mass of compound can be given by,

$Mass of compound(in grams)=Molar mass (in g)×Given mass(in g)$

Expert Solution

Answer to Problem 108AE

$305.9 mg$ Sodium lactate,

$20.0 mg CaCl2.2H2O$ ,

$29.9 mg KCl$ and

$601.2 mg NaCl$ are required to prepare lactated Ringer’s reagent.

To calculate the mass of individual elements

Explanation of Solution

Mass of Sodium = $285-315 mg$

Mass of Potassium = $14.1-17.3 mg$

Mass of Calcium = $4.9-6.0 mg$

Mass of Chlorine = $368-408 mg$

Mass of Lactate = $231-261 mg$

To calculate the mass of individual elements

Molar mass of Sodium lactate = $112.06 mg$

Molar mass of Lactate = $89.07 mg$

Molar mass of $CaCl2.2H2O$ = $147.01 mg$

Molecular mass of Calcium = $40.08 mg$

Molar mass of $KCl$ = $74.55 mg$

Molecular weight of Potassium = $39.10 mg$

Molar mass of $NaCl$ = $58.44 mg$

Molecular mass of Sodium= $22.99 mg$

The average values for each ion are,

$300.00 mg Na+,15.7 mg K+,5.45 mg Ca2+,388 mg Cl- and 246 mg Lactate$

The source of Lactate is $NaC3H5O3$

Mass of Lactate = $246 mg C3H5O3-×112.06 mg NaC3H5O389.07 mg C3H5O3-=309.5 mg$

The source of $Ca2+$ is $CaCl2.2H2O$

Mass of $CaCl2.2H2O$ = $5.45 mg Ca2+×147.01 mg CaCl2.2H2O40.08 mg Ca2+=20 mg CaCl2.2H2O$

The source of $K+$ is $KCl$

Mass of $KCl$ = $15.7 mg K+×74.55 mg KCl39.10 mg K+=29.9 mg$

Mass of $Na+$ added = $309.5 mg Sodium lactate-246.0 mg Lactate=63.5 mg Na+$

Additional amount of Sodium $236.5 mg Na+$ is added to get desired $300 mg$

Mass of Sodium added = $236.5 mg Na+×58.44 mg NaCl22.99 mg Na+=601.2 mg$

Mass of $Cl-$ added = $20.0 mg CaCl2.2H2O×70.90 mg Cl-147.01 mg CaCl2.2H2O=9.6 mg$

$20.0 mg CaCl2.2H2O =9.6 mg Cl-29.9 mg KCl-15.7 mg K+ =14.2 mg Cl-601.2 mg NaCl-236.5 mg Na+=364.7 mg Cl-$

Total $Cl-$ = $388.5 mg$

Therefore,

$305.9 mg$ Sodium lactate, $20.0 mg CaCl2.2H2O$ , $29.9 mg KCl$ and $601.2 mg NaCl$ are required to prepare lactated Ringer’s reagent.

Conclusion

The mass of individual elements was calculated using their respective molar mass and molecular weight and the given weight. A typical analytical balance can nearly measure to $0.1 mg$ . Therefore, $305.9 mg$ Sodium lactate, $20.0 mg CaCl2.2H2O$ , $29.9 mg KCl$ and $601.2 mg NaCl$ are required to prepare lactated Ringer’s reagent.

Interpretation Introduction

Interpretation:

The osmotic pressure at minimum and maximum concentration has to be calculated.

Concept Introduction:

Colligative properties of a substance include the depression in the freezing point, elevation of boiling-point and osmotic pressure. These are dependant only on the number present and not based on the solute particles present in an ideal solution.

The osmotic pressure can be given by the equation,

$Π=MRT$

$Here,Π=Osmotic pressure M=Molarity of solution R= Gas law constant T=Temperature$

Expert Solution

Answer to Problem 108AE

The range of osmotic pressure is $6.59 atm to 7.30 atm$

Explanation of Solution

To calculate the osmotic pressure at minimum and maximum concentration

Record the given info

Mass of Sodium = $285-315 mg$

Mass of Potassium = $14.1-17.3 mg$

Mass of Calcium = $4.9-6.0 mg$

Mass of Chlorine = $368-408 mg$

Mass of Lactate = $231-261 mg$

The masses of individual elements present in the reagent are recorded as shown above.

To calculate the minimum and maximum concentrations of ions

Molar mass of Lactate = $89.07 mg$

Molecular mass of Calcium = $40.08 mg$

Molecular weight of Potassium = $39.10 mg$

Molecular mass of Sodium= $22.99 mg$

At minimum concentration,

Molarity of Sodium = $285 mg Na+100. mL×1 mmol22.99 mg=0.124 M$

Molarity of Potassium = $14.1 mg K+100.mL×1 mmol39.10 mg=0.00361 M$

Molarity of Lactate = $231 mg C3H5O3-100.mL×1 mmol89.07 mg=0.0259 M$

Molarity of Calcium = $4.9 mg Ca2+100.mL×1 mmol40.08 mg=0.0012 M$

Molarity of Chlorine = $368 mg Cl-100.mL×1 mmol35.45 mg=0.104 M$

The total concentration = $0.124+0.00361+0.0012+0.104+0.0259$

= $0.259 M$

At maximum concentration,

Molarity of Sodium = $315 mg Na+100. mL×1 mmol22.99 mg=0.137 M$

Molarity of Potassium = $17.3 mg K+100.mL×1 mmol39.10 mg=0.00442 M$

Molarity of Lactate = $261 mg C3H5O3-100.mL×1 mmol89.07 mg=0.0293 M$

Molarity of Calcium = $6.0 mg Ca2+100.mL×1 mmol40.08 mg=0.0015 M$

Molarity of Chlorine = $408 mg Cl-100.mL×1 mmol35.45 mg=0.115 M$

The total concentration= $0.137+0.00442+0.0015+0.115+0.0293$

= $0.287 M$

To calculate the osmotic pressure at minimum and maximum concentration

At minimum concentration,

$π=MRT =0.259 molL×0.08206 L atmK mol×310.K =6.59 atm$

At maximum concentration,

$π=MRT =0.287 molL×0.08206 L atmK mol×310.K =7.30 atm$

At minimum concentration, osmotic pressure= $6.59 atm$

At maximum concentration, osmotic pressure= $7.30 atm$

Conclusion

The osmotic pressure at minimum and maximum concentrations was calculated using the molarities at minimum and maximum concentration. The osmotic pressure at minimum and maximum concentrations were $6.59 atm$ and $7.30 atm$ .

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Answer 5

Chapter 11, Problem 108AE

Interpretation Introduction

Interpretation:

The amount of reagents required and range of osmotic pressure has to be calculated.

Concept Introduction:

The mass of the compound is calculated by taking the products of molar mass of the compound to the given mass. The mass of compound can be given by,

$Mass of compound(in grams)=Molar mass (in g)×Given mass(in g)$

Expert Solution

Answer to Problem 108AE

$305.9 mg$ Sodium lactate,

$20.0 mg CaCl2.2H2O$ ,

$29.9 mg KCl$ and

$601.2 mg NaCl$ are required to prepare lactated Ringer’s reagent.

To calculate the mass of individual elements

Explanation of Solution

Mass of Sodium = $285-315 mg$

Mass of Potassium = $14.1-17.3 mg$

Mass of Calcium = $4.9-6.0 mg$

Mass of Chlorine = $368-408 mg$

Mass of Lactate = $231-261 mg$

To calculate the mass of individual elements

Molar mass of Sodium lactate = $112.06 mg$

Molar mass of Lactate = $89.07 mg$

Molar mass of $CaCl2.2H2O$ = $147.01 mg$

Molecular mass of Calcium = $40.08 mg$

Molar mass of $KCl$ = $74.55 mg$

Molecular weight of Potassium = $39.10 mg$

Molar mass of $NaCl$ = $58.44 mg$

Molecular mass of Sodium= $22.99 mg$

The average values for each ion are,

$300.00 mg Na+,15.7 mg K+,5.45 mg Ca2+,388 mg Cl- and 246 mg Lactate$

The source of Lactate is $NaC3H5O3$

Mass of Lactate = $246 mg C3H5O3-×112.06 mg NaC3H5O389.07 mg C3H5O3-=309.5 mg$

The source of $Ca2+$ is $CaCl2.2H2O$

Mass of $CaCl2.2H2O$ = $5.45 mg Ca2+×147.01 mg CaCl2.2H2O40.08 mg Ca2+=20 mg CaCl2.2H2O$

The source of $K+$ is $KCl$

Mass of $KCl$ = $15.7 mg K+×74.55 mg KCl39.10 mg K+=29.9 mg$

Mass of $Na+$ added = $309.5 mg Sodium lactate-246.0 mg Lactate=63.5 mg Na+$

Additional amount of Sodium $236.5 mg Na+$ is added to get desired $300 mg$

Mass of Sodium added = $236.5 mg Na+×58.44 mg NaCl22.99 mg Na+=601.2 mg$

Mass of $Cl-$ added = $20.0 mg CaCl2.2H2O×70.90 mg Cl-147.01 mg CaCl2.2H2O=9.6 mg$

$20.0 mg CaCl2.2H2O =9.6 mg Cl-29.9 mg KCl-15.7 mg K+ =14.2 mg Cl-601.2 mg NaCl-236.5 mg Na+=364.7 mg Cl-$

Total $Cl-$ = $388.5 mg$

Therefore,

$305.9 mg$ Sodium lactate, $20.0 mg CaCl2.2H2O$ , $29.9 mg KCl$ and $601.2 mg NaCl$ are required to prepare lactated Ringer’s reagent.

Conclusion

The mass of individual elements was calculated using their respective molar mass and molecular weight and the given weight. A typical analytical balance can nearly measure to $0.1 mg$ . Therefore, $305.9 mg$ Sodium lactate, $20.0 mg CaCl2.2H2O$ , $29.9 mg KCl$ and $601.2 mg NaCl$ are required to prepare lactated Ringer’s reagent.

Interpretation Introduction

Interpretation:

The osmotic pressure at minimum and maximum concentration has to be calculated.

Concept Introduction:

Colligative properties of a substance include the depression in the freezing point, elevation of boiling-point and osmotic pressure. These are dependant only on the number present and not based on the solute particles present in an ideal solution.

The osmotic pressure can be given by the equation,

$Π=MRT$

$Here,Π=Osmotic pressure M=Molarity of solution R= Gas law constant T=Temperature$

Expert Solution

Answer to Problem 108AE

The range of osmotic pressure is $6.59 atm to 7.30 atm$

Explanation of Solution

To calculate the osmotic pressure at minimum and maximum concentration

Record the given info

Mass of Sodium = $285-315 mg$

Mass of Potassium = $14.1-17.3 mg$

Mass of Calcium = $4.9-6.0 mg$

Mass of Chlorine = $368-408 mg$

Mass of Lactate = $231-261 mg$

The masses of individual elements present in the reagent are recorded as shown above.

To calculate the minimum and maximum concentrations of ions

Molar mass of Lactate = $89.07 mg$

Molecular mass of Calcium = $40.08 mg$

Molecular weight of Potassium = $39.10 mg$

Molecular mass of Sodium= $22.99 mg$

At minimum concentration,

Molarity of Sodium = $285 mg Na+100. mL×1 mmol22.99 mg=0.124 M$

Molarity of Potassium = $14.1 mg K+100.mL×1 mmol39.10 mg=0.00361 M$

Molarity of Lactate = $231 mg C3H5O3-100.mL×1 mmol89.07 mg=0.0259 M$

Molarity of Calcium = $4.9 mg Ca2+100.mL×1 mmol40.08 mg=0.0012 M$

Molarity of Chlorine = $368 mg Cl-100.mL×1 mmol35.45 mg=0.104 M$

The total concentration = $0.124+0.00361+0.0012+0.104+0.0259$

= $0.259 M$

At maximum concentration,

Molarity of Sodium = $315 mg Na+100. mL×1 mmol22.99 mg=0.137 M$

Molarity of Potassium = $17.3 mg K+100.mL×1 mmol39.10 mg=0.00442 M$

Molarity of Lactate = $261 mg C3H5O3-100.mL×1 mmol89.07 mg=0.0293 M$

Molarity of Calcium = $6.0 mg Ca2+100.mL×1 mmol40.08 mg=0.0015 M$

Molarity of Chlorine = $408 mg Cl-100.mL×1 mmol35.45 mg=0.115 M$

The total concentration= $0.137+0.00442+0.0015+0.115+0.0293$

= $0.287 M$

To calculate the osmotic pressure at minimum and maximum concentration

At minimum concentration,

$π=MRT =0.259 molL×0.08206 L atmK mol×310.K =6.59 atm$

At maximum concentration,

$π=MRT =0.287 molL×0.08206 L atmK mol×310.K =7.30 atm$

At minimum concentration, osmotic pressure= $6.59 atm$

At maximum concentration, osmotic pressure= $7.30 atm$

Conclusion

The osmotic pressure at minimum and maximum concentrations was calculated using the molarities at minimum and maximum concentration. The osmotic pressure at minimum and maximum concentrations were $6.59 atm$ and $7.30 atm$ .

Answer 6

Chapter 11, Problem 108AE

Interpretation Introduction

Interpretation:

The amount of reagents required and range of osmotic pressure has to be calculated.

Concept Introduction:

The mass of the compound is calculated by taking the products of molar mass of the compound to the given mass. The mass of compound can be given by,

$Mass of compound(in grams)=Molar mass (in g)×Given mass(in g)$

Expert Solution

Answer to Problem 108AE

$305.9 mg$ Sodium lactate,

$20.0 mg CaCl2.2H2O$ ,

$29.9 mg KCl$ and

$601.2 mg NaCl$ are required to prepare lactated Ringer’s reagent.

To calculate the mass of individual elements

Explanation of Solution

Mass of Sodium = $285-315 mg$

Mass of Potassium = $14.1-17.3 mg$

Mass of Calcium = $4.9-6.0 mg$

Mass of Chlorine = $368-408 mg$

Mass of Lactate = $231-261 mg$

To calculate the mass of individual elements

Molar mass of Sodium lactate = $112.06 mg$

Molar mass of Lactate = $89.07 mg$

Molar mass of $CaCl2.2H2O$ = $147.01 mg$

Molecular mass of Calcium = $40.08 mg$

Molar mass of $KCl$ = $74.55 mg$

Molecular weight of Potassium = $39.10 mg$

Molar mass of $NaCl$ = $58.44 mg$

Molecular mass of Sodium= $22.99 mg$

The average values for each ion are,

$300.00 mg Na+,15.7 mg K+,5.45 mg Ca2+,388 mg Cl- and 246 mg Lactate$

The source of Lactate is $NaC3H5O3$

Mass of Lactate = $246 mg C3H5O3-×112.06 mg NaC3H5O389.07 mg C3H5O3-=309.5 mg$

The source of $Ca2+$ is $CaCl2.2H2O$

Mass of $CaCl2.2H2O$ = $5.45 mg Ca2+×147.01 mg CaCl2.2H2O40.08 mg Ca2+=20 mg CaCl2.2H2O$

The source of $K+$ is $KCl$

Mass of $KCl$ = $15.7 mg K+×74.55 mg KCl39.10 mg K+=29.9 mg$

Mass of $Na+$ added = $309.5 mg Sodium lactate-246.0 mg Lactate=63.5 mg Na+$

Additional amount of Sodium $236.5 mg Na+$ is added to get desired $300 mg$

Mass of Sodium added = $236.5 mg Na+×58.44 mg NaCl22.99 mg Na+=601.2 mg$

Mass of $Cl-$ added = $20.0 mg CaCl2.2H2O×70.90 mg Cl-147.01 mg CaCl2.2H2O=9.6 mg$

$20.0 mg CaCl2.2H2O =9.6 mg Cl-29.9 mg KCl-15.7 mg K+ =14.2 mg Cl-601.2 mg NaCl-236.5 mg Na+=364.7 mg Cl-$

Total $Cl-$ = $388.5 mg$

Therefore,

$305.9 mg$ Sodium lactate, $20.0 mg CaCl2.2H2O$ , $29.9 mg KCl$ and $601.2 mg NaCl$ are required to prepare lactated Ringer’s reagent.

Conclusion

The mass of individual elements was calculated using their respective molar mass and molecular weight and the given weight. A typical analytical balance can nearly measure to $0.1 mg$ . Therefore, $305.9 mg$ Sodium lactate, $20.0 mg CaCl2.2H2O$ , $29.9 mg KCl$ and $601.2 mg NaCl$ are required to prepare lactated Ringer’s reagent.

Answer 7

Chapter 11, Problem 108AE

Interpretation Introduction

Interpretation:

The amount of reagents required and range of osmotic pressure has to be calculated.

Concept Introduction:

The mass of the compound is calculated by taking the products of molar mass of the compound to the given mass. The mass of compound can be given by,

$Mass of compound(in grams)=Molar mass (in g)×Given mass(in g)$

Answer 8

Expert Solution

Answer to Problem 108AE

$305.9 mg$ Sodium lactate,

$20.0 mg CaCl2.2H2O$ ,

$29.9 mg KCl$ and

$601.2 mg NaCl$ are required to prepare lactated Ringer’s reagent.

To calculate the mass of individual elements

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Mass of Sodium = $285-315 mg$

Mass of Potassium = $14.1-17.3 mg$

Mass of Calcium = $4.9-6.0 mg$

Mass of Chlorine = $368-408 mg$

Mass of Lactate = $231-261 mg$

To calculate the mass of individual elements

Molar mass of Sodium lactate = $112.06 mg$

Molar mass of Lactate = $89.07 mg$

Molar mass of $CaCl2.2H2O$ = $147.01 mg$

Molecular mass of Calcium = $40.08 mg$

Molar mass of $KCl$ = $74.55 mg$

Molecular weight of Potassium = $39.10 mg$

Molar mass of $NaCl$ = $58.44 mg$

Molecular mass of Sodium= $22.99 mg$

The average values for each ion are,

$300.00 mg Na+,15.7 mg K+,5.45 mg Ca2+,388 mg Cl- and 246 mg Lactate$

The source of Lactate is $NaC3H5O3$

Mass of Lactate = $246 mg C3H5O3-×112.06 mg NaC3H5O389.07 mg C3H5O3-=309.5 mg$

The source of $Ca2+$ is $CaCl2.2H2O$

Mass of $CaCl2.2H2O$ = $5.45 mg Ca2+×147.01 mg CaCl2.2H2O40.08 mg Ca2+=20 mg CaCl2.2H2O$

The source of $K+$ is $KCl$

Mass of $KCl$ = $15.7 mg K+×74.55 mg KCl39.10 mg K+=29.9 mg$

Mass of $Na+$ added = $309.5 mg Sodium lactate-246.0 mg Lactate=63.5 mg Na+$

Additional amount of Sodium $236.5 mg Na+$ is added to get desired $300 mg$

Mass of Sodium added = $236.5 mg Na+×58.44 mg NaCl22.99 mg Na+=601.2 mg$

Mass of $Cl-$ added = $20.0 mg CaCl2.2H2O×70.90 mg Cl-147.01 mg CaCl2.2H2O=9.6 mg$

$20.0 mg CaCl2.2H2O =9.6 mg Cl-29.9 mg KCl-15.7 mg K+ =14.2 mg Cl-601.2 mg NaCl-236.5 mg Na+=364.7 mg Cl-$

Total $Cl-$ = $388.5 mg$

Therefore,

$305.9 mg$ Sodium lactate, $20.0 mg CaCl2.2H2O$ , $29.9 mg KCl$ and $601.2 mg NaCl$ are required to prepare lactated Ringer’s reagent.

Answer 13

Conclusion

The mass of individual elements was calculated using their respective molar mass and molecular weight and the given weight. A typical analytical balance can nearly measure to $0.1 mg$ . Therefore, $305.9 mg$ Sodium lactate, $20.0 mg CaCl2.2H2O$ , $29.9 mg KCl$ and $601.2 mg NaCl$ are required to prepare lactated Ringer’s reagent.

Answer 14

Interpretation Introduction

Interpretation:

The osmotic pressure at minimum and maximum concentration has to be calculated.

Concept Introduction:

Colligative properties of a substance include the depression in the freezing point, elevation of boiling-point and osmotic pressure. These are dependant only on the number present and not based on the solute particles present in an ideal solution.

The osmotic pressure can be given by the equation,

$Π=MRT$

$Here,Π=Osmotic pressure M=Molarity of solution R= Gas law constant T=Temperature$

Expert Solution

Answer to Problem 108AE

The range of osmotic pressure is $6.59 atm to 7.30 atm$

Explanation of Solution

To calculate the osmotic pressure at minimum and maximum concentration

Record the given info

Mass of Sodium = $285-315 mg$

Mass of Potassium = $14.1-17.3 mg$

Mass of Calcium = $4.9-6.0 mg$

Mass of Chlorine = $368-408 mg$

Mass of Lactate = $231-261 mg$

The masses of individual elements present in the reagent are recorded as shown above.

To calculate the minimum and maximum concentrations of ions

Molar mass of Lactate = $89.07 mg$

Molecular mass of Calcium = $40.08 mg$

Molecular weight of Potassium = $39.10 mg$

Molecular mass of Sodium= $22.99 mg$

At minimum concentration,

Molarity of Sodium = $285 mg Na+100. mL×1 mmol22.99 mg=0.124 M$

Molarity of Potassium = $14.1 mg K+100.mL×1 mmol39.10 mg=0.00361 M$

Molarity of Lactate = $231 mg C3H5O3-100.mL×1 mmol89.07 mg=0.0259 M$

Molarity of Calcium = $4.9 mg Ca2+100.mL×1 mmol40.08 mg=0.0012 M$

Molarity of Chlorine = $368 mg Cl-100.mL×1 mmol35.45 mg=0.104 M$

The total concentration = $0.124+0.00361+0.0012+0.104+0.0259$

= $0.259 M$

At maximum concentration,

Molarity of Sodium = $315 mg Na+100. mL×1 mmol22.99 mg=0.137 M$

Molarity of Potassium = $17.3 mg K+100.mL×1 mmol39.10 mg=0.00442 M$

Molarity of Lactate = $261 mg C3H5O3-100.mL×1 mmol89.07 mg=0.0293 M$

Molarity of Calcium = $6.0 mg Ca2+100.mL×1 mmol40.08 mg=0.0015 M$

Molarity of Chlorine = $408 mg Cl-100.mL×1 mmol35.45 mg=0.115 M$

The total concentration= $0.137+0.00442+0.0015+0.115+0.0293$

= $0.287 M$

To calculate the osmotic pressure at minimum and maximum concentration

At minimum concentration,

$π=MRT =0.259 molL×0.08206 L atmK mol×310.K =6.59 atm$

At maximum concentration,

$π=MRT =0.287 molL×0.08206 L atmK mol×310.K =7.30 atm$

At minimum concentration, osmotic pressure= $6.59 atm$

At maximum concentration, osmotic pressure= $7.30 atm$

Conclusion

The osmotic pressure at minimum and maximum concentrations was calculated using the molarities at minimum and maximum concentration. The osmotic pressure at minimum and maximum concentrations were $6.59 atm$ and $7.30 atm$ .

Answer 15

Interpretation Introduction

Interpretation:

The osmotic pressure at minimum and maximum concentration has to be calculated.

Concept Introduction:

Colligative properties of a substance include the depression in the freezing point, elevation of boiling-point and osmotic pressure. These are dependant only on the number present and not based on the solute particles present in an ideal solution.

The osmotic pressure can be given by the equation,

$Π=MRT$

$Here,Π=Osmotic pressure M=Molarity of solution R= Gas law constant T=Temperature$

Answer 16

Expert Solution

Answer to Problem 108AE

The range of osmotic pressure is $6.59 atm to 7.30 atm$

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

To calculate the osmotic pressure at minimum and maximum concentration

Record the given info

Mass of Sodium = $285-315 mg$

Mass of Potassium = $14.1-17.3 mg$

Mass of Calcium = $4.9-6.0 mg$

Mass of Chlorine = $368-408 mg$

Mass of Lactate = $231-261 mg$

The masses of individual elements present in the reagent are recorded as shown above.

To calculate the minimum and maximum concentrations of ions

Molar mass of Lactate = $89.07 mg$

Molecular mass of Calcium = $40.08 mg$

Molecular weight of Potassium = $39.10 mg$

Molecular mass of Sodium= $22.99 mg$

At minimum concentration,

Molarity of Sodium = $285 mg Na+100. mL×1 mmol22.99 mg=0.124 M$

Molarity of Potassium = $14.1 mg K+100.mL×1 mmol39.10 mg=0.00361 M$

Molarity of Lactate = $231 mg C3H5O3-100.mL×1 mmol89.07 mg=0.0259 M$

Molarity of Calcium = $4.9 mg Ca2+100.mL×1 mmol40.08 mg=0.0012 M$

Molarity of Chlorine = $368 mg Cl-100.mL×1 mmol35.45 mg=0.104 M$

The total concentration = $0.124+0.00361+0.0012+0.104+0.0259$

= $0.259 M$

At maximum concentration,

Molarity of Sodium = $315 mg Na+100. mL×1 mmol22.99 mg=0.137 M$

Molarity of Potassium = $17.3 mg K+100.mL×1 mmol39.10 mg=0.00442 M$

Molarity of Lactate = $261 mg C3H5O3-100.mL×1 mmol89.07 mg=0.0293 M$

Molarity of Calcium = $6.0 mg Ca2+100.mL×1 mmol40.08 mg=0.0015 M$

Molarity of Chlorine = $408 mg Cl-100.mL×1 mmol35.45 mg=0.115 M$

The total concentration= $0.137+0.00442+0.0015+0.115+0.0293$

= $0.287 M$

To calculate the osmotic pressure at minimum and maximum concentration

At minimum concentration,

$π=MRT =0.259 molL×0.08206 L atmK mol×310.K =6.59 atm$

At maximum concentration,

$π=MRT =0.287 molL×0.08206 L atmK mol×310.K =7.30 atm$

At minimum concentration, osmotic pressure= $6.59 atm$

At maximum concentration, osmotic pressure= $7.30 atm$

Answer 21

Conclusion

The osmotic pressure at minimum and maximum concentrations was calculated using the molarities at minimum and maximum concentration. The osmotic pressure at minimum and maximum concentrations were $6.59 atm$ and $7.30 atm$ .