The reason as to why people who live at high altitudes are advised to add salt to water when boiling food like pasta is to be explained. The mole fraction of NaCl needed to raise the boiling point of H 2 O by 3 ° C is to be calculated. Whether the given amount of salt added to water substantially changes the boiling point or not is to be identified. Concept introduction: When a non-volatile solute is added to a solvent, the boiling point of the solution gets raised in comparison to that of the pure solvent. The formula for elevation in boiling point is given as, Δ T b = m K b Where, • m represents the molality of the solute. • K b represents the boiling point elevation constant.

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Answer 1

Question

Chapter 7, Problem 7.71E

Interpretation Introduction

Interpretation:

The reason as to why people who live at high altitudes are advised to add salt to water when boiling food like pasta is to be explained. The mole fraction of $NaCl$ needed to raise the boiling point of $H2O$ by $3°C$ is to be calculated. Whether the given amount of salt added to water substantially changes the boiling point or not is to be identified.

Concept introduction:

When a non-volatile solute is added to a solvent, the boiling point of the solution gets raised in comparison to that of the pure solvent. The formula for elevation in boiling point is given as,

$ΔTb=mKb$

Where,

• $m$ represents the molality of the solute.

• $Kb$ represents the boiling point elevation constant.

Expert Solution & Answer

Answer to Problem 7.71E

People who live at high altitudes are advised to add salt to water when boiling food like pasta because the pressure at high altitudes is low and the water starts boiling below its normal boiling point. Therefore, salt is added to increase the boiling point of the water. The mole fraction of $NaCl$ needed to raise the boiling point of $H2O$ by $3°C$ is $0.0957$ .The given amount of salt added to water does not substantially change the boiling point.

Explanation of Solution

The pressure at high altitudes is low and the water starts boiling below its normal boiling point. Therefore, the water starts boiling before the food cooks and it is advised to add salt to water when boiling food like pasta.

The boiling point elevation constant of water is $0.51°C/molal$ .

The required elevation in the boiling temperature of the water is $3°C$ .

The formula for elevation in boiling point is given as,

$ΔTb=mKb ⋅⋅⋅(1)$

Where,

• $m$ represents the molality of the solute.

• $Kb$ represents the boiling point elevation constant.

Rearrange the above equation for the value of $m$ .

Substitute the value of $ΔTb$ and $Kb$ in the above equation.

$m=3°C0.51°C/molal=(5.8834 molal)(1 mol/kg1 molal)=5.8824 mol/kg$

The molality of the salt solution is $5.8824 mol/kg$ . Therefore, $5.8824 mol$ of the $NaCl$ is dissolved in $1 kg$ of water.

The molar mass of the water is $18 g/mol$ .

The number of mole of a substance is given as:

$n=MMm$

Where,

• $M$ represents the mass of the substance.

• $Mm$ represents the molar mass of the substance.

Substitute the value of the mass and molar mass of water in the above equation.

$n=(1 kg)(1000 g1 kg)18 g/mol=55.55 mol$

The number of moles of water is $55.55 mol$ .

The mole fraction of a substance present in two-component system is given as,

$xa=nana+nb$

Where,

• $na$ represents the number of moles of substance $a$ .

• $nb$ represents the number of moles of substance $b$ .

Substitute the value of a number of moles of water and the number of moles of sodium chloride in the above equation for the value of mole fraction of sodium chloride.

$xa=5.8824 mol5.8824 mol+55.5555 mol=5.8824 mol61.4379 mol=0.0957$

The mole fraction of $NaCl$ is needed to raise the boiling point of $H2O$ by $3°C$ is $0.0957$ .

One teaspoon of salt is dissolved into four quarts of water.

The molar mass of sodium chloride is $58.44 g/mol$ .

The standard amount present in one teaspoon is $5 mL$ .

One quart has $946.353 mL$ .

The density of water is $1 g/mL$ .

The density of sodium chloride is $2.16 g/mL$ .

The relation between mass, volume and density is given as,

$M=DV$ (2)

Where,

• $M$ represents the mass of the substance.

• $D$ represents the density of the substance.

• $V$ represents the volume of the substance.

Substitute the value of density and volume of water in the equation (2).

$M=(1 g/mL)(4 quarts)(946.353 mL1 quart)=3785.412 g$

Substitute the value of density and volume of sodium chloride in the equation (2).

$M=(2.16 g/mL)(1 teaspoon)(5 mL1 teaspoon)=10.8 g$

The molality of a salt solution is given as,

$m=MsMmMW$

Where,

• $Ms$ represents the mass of the solute.

• $Mm$ represents the molar mass of the solute.

• $Mw$ represents the mass of the solvent.

Substitute the molar mass of sodium chloride, the mass of the sodium chloride and mass of the water in the above equation.

$m=10.8 g(58.44 g/mol)(3785.412 g)(1 kg1000 g)=(0.0488 mol/kg)(1 molal1 mol/kg )=0.0488 molal$

Therefore, the molality of the solution that is formed by dissolving one teaspoon of salt into four quarts of water is $0.0488 molal$ .

Substitute the value of molality of salt solution and $Kb$ in the equation (1).

$ΔTb=(0.0488 molal)(0.51°C/molal)=0.0249°C$

The elevation in the boiling point of water is very low. Therefore, the given amount of salt added to water does not substantially change the boiling point.

Conclusion

People who live at high altitudes are advised to add salt to water when boiling food like pasta to increase the boiling point of water The mole fraction of $NaCl$ is needed to raise the boiling point of $H2O$ by $3°C$ is $0.0957$ .The given amount of salt added to water does not substantially change the boiling point.

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Answer 2

Question

Chapter 7, Problem 7.71E

Interpretation Introduction

Interpretation:

The reason as to why people who live at high altitudes are advised to add salt to water when boiling food like pasta is to be explained. The mole fraction of $NaCl$ needed to raise the boiling point of $H2O$ by $3°C$ is to be calculated. Whether the given amount of salt added to water substantially changes the boiling point or not is to be identified.

Concept introduction:

When a non-volatile solute is added to a solvent, the boiling point of the solution gets raised in comparison to that of the pure solvent. The formula for elevation in boiling point is given as,

$ΔTb=mKb$

Where,

• $m$ represents the molality of the solute.

• $Kb$ represents the boiling point elevation constant.

Expert Solution & Answer

Answer to Problem 7.71E

People who live at high altitudes are advised to add salt to water when boiling food like pasta because the pressure at high altitudes is low and the water starts boiling below its normal boiling point. Therefore, salt is added to increase the boiling point of the water. The mole fraction of $NaCl$ needed to raise the boiling point of $H2O$ by $3°C$ is $0.0957$ .The given amount of salt added to water does not substantially change the boiling point.

Explanation of Solution

The pressure at high altitudes is low and the water starts boiling below its normal boiling point. Therefore, the water starts boiling before the food cooks and it is advised to add salt to water when boiling food like pasta.

The boiling point elevation constant of water is $0.51°C/molal$ .

The required elevation in the boiling temperature of the water is $3°C$ .

The formula for elevation in boiling point is given as,

$ΔTb=mKb ⋅⋅⋅(1)$

Where,

• $m$ represents the molality of the solute.

• $Kb$ represents the boiling point elevation constant.

Rearrange the above equation for the value of $m$ .

Substitute the value of $ΔTb$ and $Kb$ in the above equation.

$m=3°C0.51°C/molal=(5.8834 molal)(1 mol/kg1 molal)=5.8824 mol/kg$

The molality of the salt solution is $5.8824 mol/kg$ . Therefore, $5.8824 mol$ of the $NaCl$ is dissolved in $1 kg$ of water.

The molar mass of the water is $18 g/mol$ .

The number of mole of a substance is given as:

$n=MMm$

Where,

• $M$ represents the mass of the substance.

• $Mm$ represents the molar mass of the substance.

Substitute the value of the mass and molar mass of water in the above equation.

$n=(1 kg)(1000 g1 kg)18 g/mol=55.55 mol$

The number of moles of water is $55.55 mol$ .

The mole fraction of a substance present in two-component system is given as,

$xa=nana+nb$

Where,

• $na$ represents the number of moles of substance $a$ .

• $nb$ represents the number of moles of substance $b$ .

Substitute the value of a number of moles of water and the number of moles of sodium chloride in the above equation for the value of mole fraction of sodium chloride.

$xa=5.8824 mol5.8824 mol+55.5555 mol=5.8824 mol61.4379 mol=0.0957$

The mole fraction of $NaCl$ is needed to raise the boiling point of $H2O$ by $3°C$ is $0.0957$ .

One teaspoon of salt is dissolved into four quarts of water.

The molar mass of sodium chloride is $58.44 g/mol$ .

The standard amount present in one teaspoon is $5 mL$ .

One quart has $946.353 mL$ .

The density of water is $1 g/mL$ .

The density of sodium chloride is $2.16 g/mL$ .

The relation between mass, volume and density is given as,

$M=DV$ (2)

Where,

• $M$ represents the mass of the substance.

• $D$ represents the density of the substance.

• $V$ represents the volume of the substance.

Substitute the value of density and volume of water in the equation (2).

$M=(1 g/mL)(4 quarts)(946.353 mL1 quart)=3785.412 g$

Substitute the value of density and volume of sodium chloride in the equation (2).

$M=(2.16 g/mL)(1 teaspoon)(5 mL1 teaspoon)=10.8 g$

The molality of a salt solution is given as,

$m=MsMmMW$

Where,

• $Ms$ represents the mass of the solute.

• $Mm$ represents the molar mass of the solute.

• $Mw$ represents the mass of the solvent.

Substitute the molar mass of sodium chloride, the mass of the sodium chloride and mass of the water in the above equation.

$m=10.8 g(58.44 g/mol)(3785.412 g)(1 kg1000 g)=(0.0488 mol/kg)(1 molal1 mol/kg )=0.0488 molal$

Therefore, the molality of the solution that is formed by dissolving one teaspoon of salt into four quarts of water is $0.0488 molal$ .

Substitute the value of molality of salt solution and $Kb$ in the equation (1).

$ΔTb=(0.0488 molal)(0.51°C/molal)=0.0249°C$

The elevation in the boiling point of water is very low. Therefore, the given amount of salt added to water does not substantially change the boiling point.

Conclusion

People who live at high altitudes are advised to add salt to water when boiling food like pasta to increase the boiling point of water The mole fraction of $NaCl$ is needed to raise the boiling point of $H2O$ by $3°C$ is $0.0957$ .The given amount of salt added to water does not substantially change the boiling point.

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Answer 3

Question

Chapter 7, Problem 7.71E

Interpretation Introduction

Interpretation:

The reason as to why people who live at high altitudes are advised to add salt to water when boiling food like pasta is to be explained. The mole fraction of $NaCl$ needed to raise the boiling point of $H2O$ by $3°C$ is to be calculated. Whether the given amount of salt added to water substantially changes the boiling point or not is to be identified.

Concept introduction:

When a non-volatile solute is added to a solvent, the boiling point of the solution gets raised in comparison to that of the pure solvent. The formula for elevation in boiling point is given as,

$ΔTb=mKb$

Where,

• $m$ represents the molality of the solute.

• $Kb$ represents the boiling point elevation constant.

Expert Solution & Answer

Answer to Problem 7.71E

People who live at high altitudes are advised to add salt to water when boiling food like pasta because the pressure at high altitudes is low and the water starts boiling below its normal boiling point. Therefore, salt is added to increase the boiling point of the water. The mole fraction of $NaCl$ needed to raise the boiling point of $H2O$ by $3°C$ is $0.0957$ .The given amount of salt added to water does not substantially change the boiling point.

Explanation of Solution

The pressure at high altitudes is low and the water starts boiling below its normal boiling point. Therefore, the water starts boiling before the food cooks and it is advised to add salt to water when boiling food like pasta.

The boiling point elevation constant of water is $0.51°C/molal$ .

The required elevation in the boiling temperature of the water is $3°C$ .

The formula for elevation in boiling point is given as,

$ΔTb=mKb ⋅⋅⋅(1)$

Where,

• $m$ represents the molality of the solute.

• $Kb$ represents the boiling point elevation constant.

Rearrange the above equation for the value of $m$ .

Substitute the value of $ΔTb$ and $Kb$ in the above equation.

$m=3°C0.51°C/molal=(5.8834 molal)(1 mol/kg1 molal)=5.8824 mol/kg$

The molality of the salt solution is $5.8824 mol/kg$ . Therefore, $5.8824 mol$ of the $NaCl$ is dissolved in $1 kg$ of water.

The molar mass of the water is $18 g/mol$ .

The number of mole of a substance is given as:

$n=MMm$

Where,

• $M$ represents the mass of the substance.

• $Mm$ represents the molar mass of the substance.

Substitute the value of the mass and molar mass of water in the above equation.

$n=(1 kg)(1000 g1 kg)18 g/mol=55.55 mol$

The number of moles of water is $55.55 mol$ .

The mole fraction of a substance present in two-component system is given as,

$xa=nana+nb$

Where,

• $na$ represents the number of moles of substance $a$ .

• $nb$ represents the number of moles of substance $b$ .

Substitute the value of a number of moles of water and the number of moles of sodium chloride in the above equation for the value of mole fraction of sodium chloride.

$xa=5.8824 mol5.8824 mol+55.5555 mol=5.8824 mol61.4379 mol=0.0957$

The mole fraction of $NaCl$ is needed to raise the boiling point of $H2O$ by $3°C$ is $0.0957$ .

One teaspoon of salt is dissolved into four quarts of water.

The molar mass of sodium chloride is $58.44 g/mol$ .

The standard amount present in one teaspoon is $5 mL$ .

One quart has $946.353 mL$ .

The density of water is $1 g/mL$ .

The density of sodium chloride is $2.16 g/mL$ .

The relation between mass, volume and density is given as,

$M=DV$ (2)

Where,

• $M$ represents the mass of the substance.

• $D$ represents the density of the substance.

• $V$ represents the volume of the substance.

Substitute the value of density and volume of water in the equation (2).

$M=(1 g/mL)(4 quarts)(946.353 mL1 quart)=3785.412 g$

Substitute the value of density and volume of sodium chloride in the equation (2).

$M=(2.16 g/mL)(1 teaspoon)(5 mL1 teaspoon)=10.8 g$

The molality of a salt solution is given as,

$m=MsMmMW$

Where,

• $Ms$ represents the mass of the solute.

• $Mm$ represents the molar mass of the solute.

• $Mw$ represents the mass of the solvent.

Substitute the molar mass of sodium chloride, the mass of the sodium chloride and mass of the water in the above equation.

$m=10.8 g(58.44 g/mol)(3785.412 g)(1 kg1000 g)=(0.0488 mol/kg)(1 molal1 mol/kg )=0.0488 molal$

Therefore, the molality of the solution that is formed by dissolving one teaspoon of salt into four quarts of water is $0.0488 molal$ .

Substitute the value of molality of salt solution and $Kb$ in the equation (1).

$ΔTb=(0.0488 molal)(0.51°C/molal)=0.0249°C$

The elevation in the boiling point of water is very low. Therefore, the given amount of salt added to water does not substantially change the boiling point.

Conclusion

People who live at high altitudes are advised to add salt to water when boiling food like pasta to increase the boiling point of water The mole fraction of $NaCl$ is needed to raise the boiling point of $H2O$ by $3°C$ is $0.0957$ .The given amount of salt added to water does not substantially change the boiling point.

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Answer 4

Question

Chapter 7, Problem 7.71E

Interpretation Introduction

Interpretation:

The reason as to why people who live at high altitudes are advised to add salt to water when boiling food like pasta is to be explained. The mole fraction of $NaCl$ needed to raise the boiling point of $H2O$ by $3°C$ is to be calculated. Whether the given amount of salt added to water substantially changes the boiling point or not is to be identified.

Concept introduction:

When a non-volatile solute is added to a solvent, the boiling point of the solution gets raised in comparison to that of the pure solvent. The formula for elevation in boiling point is given as,

$ΔTb=mKb$

Where,

• $m$ represents the molality of the solute.

• $Kb$ represents the boiling point elevation constant.

Expert Solution & Answer

Answer to Problem 7.71E

People who live at high altitudes are advised to add salt to water when boiling food like pasta because the pressure at high altitudes is low and the water starts boiling below its normal boiling point. Therefore, salt is added to increase the boiling point of the water. The mole fraction of $NaCl$ needed to raise the boiling point of $H2O$ by $3°C$ is $0.0957$ .The given amount of salt added to water does not substantially change the boiling point.

Explanation of Solution

The pressure at high altitudes is low and the water starts boiling below its normal boiling point. Therefore, the water starts boiling before the food cooks and it is advised to add salt to water when boiling food like pasta.

The boiling point elevation constant of water is $0.51°C/molal$ .

The required elevation in the boiling temperature of the water is $3°C$ .

The formula for elevation in boiling point is given as,

$ΔTb=mKb ⋅⋅⋅(1)$

Where,

• $m$ represents the molality of the solute.

• $Kb$ represents the boiling point elevation constant.

Rearrange the above equation for the value of $m$ .

Substitute the value of $ΔTb$ and $Kb$ in the above equation.

$m=3°C0.51°C/molal=(5.8834 molal)(1 mol/kg1 molal)=5.8824 mol/kg$

The molality of the salt solution is $5.8824 mol/kg$ . Therefore, $5.8824 mol$ of the $NaCl$ is dissolved in $1 kg$ of water.

The molar mass of the water is $18 g/mol$ .

The number of mole of a substance is given as:

$n=MMm$

Where,

• $M$ represents the mass of the substance.

• $Mm$ represents the molar mass of the substance.

Substitute the value of the mass and molar mass of water in the above equation.

$n=(1 kg)(1000 g1 kg)18 g/mol=55.55 mol$

The number of moles of water is $55.55 mol$ .

The mole fraction of a substance present in two-component system is given as,

$xa=nana+nb$

Where,

• $na$ represents the number of moles of substance $a$ .

• $nb$ represents the number of moles of substance $b$ .

Substitute the value of a number of moles of water and the number of moles of sodium chloride in the above equation for the value of mole fraction of sodium chloride.

$xa=5.8824 mol5.8824 mol+55.5555 mol=5.8824 mol61.4379 mol=0.0957$

The mole fraction of $NaCl$ is needed to raise the boiling point of $H2O$ by $3°C$ is $0.0957$ .

One teaspoon of salt is dissolved into four quarts of water.

The molar mass of sodium chloride is $58.44 g/mol$ .

The standard amount present in one teaspoon is $5 mL$ .

One quart has $946.353 mL$ .

The density of water is $1 g/mL$ .

The density of sodium chloride is $2.16 g/mL$ .

The relation between mass, volume and density is given as,

$M=DV$ (2)

Where,

• $M$ represents the mass of the substance.

• $D$ represents the density of the substance.

• $V$ represents the volume of the substance.

Substitute the value of density and volume of water in the equation (2).

$M=(1 g/mL)(4 quarts)(946.353 mL1 quart)=3785.412 g$

Substitute the value of density and volume of sodium chloride in the equation (2).

$M=(2.16 g/mL)(1 teaspoon)(5 mL1 teaspoon)=10.8 g$

The molality of a salt solution is given as,

$m=MsMmMW$

Where,

• $Ms$ represents the mass of the solute.

• $Mm$ represents the molar mass of the solute.

• $Mw$ represents the mass of the solvent.

Substitute the molar mass of sodium chloride, the mass of the sodium chloride and mass of the water in the above equation.

$m=10.8 g(58.44 g/mol)(3785.412 g)(1 kg1000 g)=(0.0488 mol/kg)(1 molal1 mol/kg )=0.0488 molal$

Therefore, the molality of the solution that is formed by dissolving one teaspoon of salt into four quarts of water is $0.0488 molal$ .

Substitute the value of molality of salt solution and $Kb$ in the equation (1).

$ΔTb=(0.0488 molal)(0.51°C/molal)=0.0249°C$

The elevation in the boiling point of water is very low. Therefore, the given amount of salt added to water does not substantially change the boiling point.

Conclusion

People who live at high altitudes are advised to add salt to water when boiling food like pasta to increase the boiling point of water The mole fraction of $NaCl$ is needed to raise the boiling point of $H2O$ by $3°C$ is $0.0957$ .The given amount of salt added to water does not substantially change the boiling point.

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Answer 5

Chapter 7, Problem 7.71E

Interpretation Introduction

Interpretation:

The reason as to why people who live at high altitudes are advised to add salt to water when boiling food like pasta is to be explained. The mole fraction of $NaCl$ needed to raise the boiling point of $H2O$ by $3°C$ is to be calculated. Whether the given amount of salt added to water substantially changes the boiling point or not is to be identified.

Concept introduction:

When a non-volatile solute is added to a solvent, the boiling point of the solution gets raised in comparison to that of the pure solvent. The formula for elevation in boiling point is given as,

$ΔTb=mKb$

Where,

• $m$ represents the molality of the solute.

• $Kb$ represents the boiling point elevation constant.

Expert Solution & Answer

Answer to Problem 7.71E

People who live at high altitudes are advised to add salt to water when boiling food like pasta because the pressure at high altitudes is low and the water starts boiling below its normal boiling point. Therefore, salt is added to increase the boiling point of the water. The mole fraction of $NaCl$ needed to raise the boiling point of $H2O$ by $3°C$ is $0.0957$ .The given amount of salt added to water does not substantially change the boiling point.

Explanation of Solution

The pressure at high altitudes is low and the water starts boiling below its normal boiling point. Therefore, the water starts boiling before the food cooks and it is advised to add salt to water when boiling food like pasta.

The boiling point elevation constant of water is $0.51°C/molal$ .

The required elevation in the boiling temperature of the water is $3°C$ .

The formula for elevation in boiling point is given as,

$ΔTb=mKb ⋅⋅⋅(1)$

Where,

• $m$ represents the molality of the solute.

• $Kb$ represents the boiling point elevation constant.

Rearrange the above equation for the value of $m$ .

Substitute the value of $ΔTb$ and $Kb$ in the above equation.

$m=3°C0.51°C/molal=(5.8834 molal)(1 mol/kg1 molal)=5.8824 mol/kg$

The molality of the salt solution is $5.8824 mol/kg$ . Therefore, $5.8824 mol$ of the $NaCl$ is dissolved in $1 kg$ of water.

The molar mass of the water is $18 g/mol$ .

The number of mole of a substance is given as:

$n=MMm$

Where,

• $M$ represents the mass of the substance.

• $Mm$ represents the molar mass of the substance.

Substitute the value of the mass and molar mass of water in the above equation.

$n=(1 kg)(1000 g1 kg)18 g/mol=55.55 mol$

The number of moles of water is $55.55 mol$ .

The mole fraction of a substance present in two-component system is given as,

$xa=nana+nb$

Where,

• $na$ represents the number of moles of substance $a$ .

• $nb$ represents the number of moles of substance $b$ .

Substitute the value of a number of moles of water and the number of moles of sodium chloride in the above equation for the value of mole fraction of sodium chloride.

$xa=5.8824 mol5.8824 mol+55.5555 mol=5.8824 mol61.4379 mol=0.0957$

The mole fraction of $NaCl$ is needed to raise the boiling point of $H2O$ by $3°C$ is $0.0957$ .

One teaspoon of salt is dissolved into four quarts of water.

The molar mass of sodium chloride is $58.44 g/mol$ .

The standard amount present in one teaspoon is $5 mL$ .

One quart has $946.353 mL$ .

The density of water is $1 g/mL$ .

The density of sodium chloride is $2.16 g/mL$ .

The relation between mass, volume and density is given as,

$M=DV$ (2)

Where,

• $M$ represents the mass of the substance.

• $D$ represents the density of the substance.

• $V$ represents the volume of the substance.

Substitute the value of density and volume of water in the equation (2).

$M=(1 g/mL)(4 quarts)(946.353 mL1 quart)=3785.412 g$

Substitute the value of density and volume of sodium chloride in the equation (2).

$M=(2.16 g/mL)(1 teaspoon)(5 mL1 teaspoon)=10.8 g$

The molality of a salt solution is given as,

$m=MsMmMW$

Where,

• $Ms$ represents the mass of the solute.

• $Mm$ represents the molar mass of the solute.

• $Mw$ represents the mass of the solvent.

Substitute the molar mass of sodium chloride, the mass of the sodium chloride and mass of the water in the above equation.

$m=10.8 g(58.44 g/mol)(3785.412 g)(1 kg1000 g)=(0.0488 mol/kg)(1 molal1 mol/kg )=0.0488 molal$

Therefore, the molality of the solution that is formed by dissolving one teaspoon of salt into four quarts of water is $0.0488 molal$ .

Substitute the value of molality of salt solution and $Kb$ in the equation (1).

$ΔTb=(0.0488 molal)(0.51°C/molal)=0.0249°C$

The elevation in the boiling point of water is very low. Therefore, the given amount of salt added to water does not substantially change the boiling point.

Conclusion

People who live at high altitudes are advised to add salt to water when boiling food like pasta to increase the boiling point of water The mole fraction of $NaCl$ is needed to raise the boiling point of $H2O$ by $3°C$ is $0.0957$ .The given amount of salt added to water does not substantially change the boiling point.

Answer 6

Chapter 7, Problem 7.71E

Interpretation Introduction

Interpretation:

The reason as to why people who live at high altitudes are advised to add salt to water when boiling food like pasta is to be explained. The mole fraction of $NaCl$ needed to raise the boiling point of $H2O$ by $3°C$ is to be calculated. Whether the given amount of salt added to water substantially changes the boiling point or not is to be identified.

Concept introduction:

When a non-volatile solute is added to a solvent, the boiling point of the solution gets raised in comparison to that of the pure solvent. The formula for elevation in boiling point is given as,

$ΔTb=mKb$

Where,

• $m$ represents the molality of the solute.

• $Kb$ represents the boiling point elevation constant.

Expert Solution & Answer

Answer to Problem 7.71E

People who live at high altitudes are advised to add salt to water when boiling food like pasta because the pressure at high altitudes is low and the water starts boiling below its normal boiling point. Therefore, salt is added to increase the boiling point of the water. The mole fraction of $NaCl$ needed to raise the boiling point of $H2O$ by $3°C$ is $0.0957$ .The given amount of salt added to water does not substantially change the boiling point.

Explanation of Solution

The pressure at high altitudes is low and the water starts boiling below its normal boiling point. Therefore, the water starts boiling before the food cooks and it is advised to add salt to water when boiling food like pasta.

The boiling point elevation constant of water is $0.51°C/molal$ .

The required elevation in the boiling temperature of the water is $3°C$ .

The formula for elevation in boiling point is given as,

$ΔTb=mKb ⋅⋅⋅(1)$

Where,

• $m$ represents the molality of the solute.

• $Kb$ represents the boiling point elevation constant.

Rearrange the above equation for the value of $m$ .

Substitute the value of $ΔTb$ and $Kb$ in the above equation.

$m=3°C0.51°C/molal=(5.8834 molal)(1 mol/kg1 molal)=5.8824 mol/kg$

The molality of the salt solution is $5.8824 mol/kg$ . Therefore, $5.8824 mol$ of the $NaCl$ is dissolved in $1 kg$ of water.

The molar mass of the water is $18 g/mol$ .

The number of mole of a substance is given as:

$n=MMm$

Where,

• $M$ represents the mass of the substance.

• $Mm$ represents the molar mass of the substance.

Substitute the value of the mass and molar mass of water in the above equation.

$n=(1 kg)(1000 g1 kg)18 g/mol=55.55 mol$

The number of moles of water is $55.55 mol$ .

The mole fraction of a substance present in two-component system is given as,

$xa=nana+nb$

Where,

• $na$ represents the number of moles of substance $a$ .

• $nb$ represents the number of moles of substance $b$ .

Substitute the value of a number of moles of water and the number of moles of sodium chloride in the above equation for the value of mole fraction of sodium chloride.

$xa=5.8824 mol5.8824 mol+55.5555 mol=5.8824 mol61.4379 mol=0.0957$

The mole fraction of $NaCl$ is needed to raise the boiling point of $H2O$ by $3°C$ is $0.0957$ .

One teaspoon of salt is dissolved into four quarts of water.

The molar mass of sodium chloride is $58.44 g/mol$ .

The standard amount present in one teaspoon is $5 mL$ .

One quart has $946.353 mL$ .

The density of water is $1 g/mL$ .

The density of sodium chloride is $2.16 g/mL$ .

The relation between mass, volume and density is given as,

$M=DV$ (2)

Where,

• $M$ represents the mass of the substance.

• $D$ represents the density of the substance.

• $V$ represents the volume of the substance.

Substitute the value of density and volume of water in the equation (2).

$M=(1 g/mL)(4 quarts)(946.353 mL1 quart)=3785.412 g$

Substitute the value of density and volume of sodium chloride in the equation (2).

$M=(2.16 g/mL)(1 teaspoon)(5 mL1 teaspoon)=10.8 g$

The molality of a salt solution is given as,

$m=MsMmMW$

Where,

• $Ms$ represents the mass of the solute.

• $Mm$ represents the molar mass of the solute.

• $Mw$ represents the mass of the solvent.

Substitute the molar mass of sodium chloride, the mass of the sodium chloride and mass of the water in the above equation.

$m=10.8 g(58.44 g/mol)(3785.412 g)(1 kg1000 g)=(0.0488 mol/kg)(1 molal1 mol/kg )=0.0488 molal$

Therefore, the molality of the solution that is formed by dissolving one teaspoon of salt into four quarts of water is $0.0488 molal$ .

Substitute the value of molality of salt solution and $Kb$ in the equation (1).

$ΔTb=(0.0488 molal)(0.51°C/molal)=0.0249°C$

The elevation in the boiling point of water is very low. Therefore, the given amount of salt added to water does not substantially change the boiling point.

Conclusion

People who live at high altitudes are advised to add salt to water when boiling food like pasta to increase the boiling point of water The mole fraction of $NaCl$ is needed to raise the boiling point of $H2O$ by $3°C$ is $0.0957$ .The given amount of salt added to water does not substantially change the boiling point.

Answer 7

Chapter 7, Problem 7.71E

Interpretation Introduction

Interpretation:

The reason as to why people who live at high altitudes are advised to add salt to water when boiling food like pasta is to be explained. The mole fraction of $NaCl$ needed to raise the boiling point of $H2O$ by $3°C$ is to be calculated. Whether the given amount of salt added to water substantially changes the boiling point or not is to be identified.

Concept introduction:

When a non-volatile solute is added to a solvent, the boiling point of the solution gets raised in comparison to that of the pure solvent. The formula for elevation in boiling point is given as,

$ΔTb=mKb$

Where,

• $m$ represents the molality of the solute.

• $Kb$ represents the boiling point elevation constant.

Answer 8

Expert Solution & Answer

Answer to Problem 7.71E

People who live at high altitudes are advised to add salt to water when boiling food like pasta because the pressure at high altitudes is low and the water starts boiling below its normal boiling point. Therefore, salt is added to increase the boiling point of the water. The mole fraction of $NaCl$ needed to raise the boiling point of $H2O$ by $3°C$ is $0.0957$ .The given amount of salt added to water does not substantially change the boiling point.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The pressure at high altitudes is low and the water starts boiling below its normal boiling point. Therefore, the water starts boiling before the food cooks and it is advised to add salt to water when boiling food like pasta.

The boiling point elevation constant of water is $0.51°C/molal$ .

The required elevation in the boiling temperature of the water is $3°C$ .

The formula for elevation in boiling point is given as,

$ΔTb=mKb ⋅⋅⋅(1)$

Where,

• $m$ represents the molality of the solute.

• $Kb$ represents the boiling point elevation constant.

Rearrange the above equation for the value of $m$ .

Substitute the value of $ΔTb$ and $Kb$ in the above equation.

$m=3°C0.51°C/molal=(5.8834 molal)(1 mol/kg1 molal)=5.8824 mol/kg$

The molality of the salt solution is $5.8824 mol/kg$ . Therefore, $5.8824 mol$ of the $NaCl$ is dissolved in $1 kg$ of water.

The molar mass of the water is $18 g/mol$ .

The number of mole of a substance is given as:

$n=MMm$

Where,

• $M$ represents the mass of the substance.

• $Mm$ represents the molar mass of the substance.

Substitute the value of the mass and molar mass of water in the above equation.

$n=(1 kg)(1000 g1 kg)18 g/mol=55.55 mol$

The number of moles of water is $55.55 mol$ .

The mole fraction of a substance present in two-component system is given as,

$xa=nana+nb$

Where,

• $na$ represents the number of moles of substance $a$ .

• $nb$ represents the number of moles of substance $b$ .

Substitute the value of a number of moles of water and the number of moles of sodium chloride in the above equation for the value of mole fraction of sodium chloride.

$xa=5.8824 mol5.8824 mol+55.5555 mol=5.8824 mol61.4379 mol=0.0957$

The mole fraction of $NaCl$ is needed to raise the boiling point of $H2O$ by $3°C$ is $0.0957$ .

One teaspoon of salt is dissolved into four quarts of water.

The molar mass of sodium chloride is $58.44 g/mol$ .

The standard amount present in one teaspoon is $5 mL$ .

One quart has $946.353 mL$ .

The density of water is $1 g/mL$ .

The density of sodium chloride is $2.16 g/mL$ .

The relation between mass, volume and density is given as,

$M=DV$ (2)

Where,

• $M$ represents the mass of the substance.

• $D$ represents the density of the substance.

• $V$ represents the volume of the substance.

Substitute the value of density and volume of water in the equation (2).

$M=(1 g/mL)(4 quarts)(946.353 mL1 quart)=3785.412 g$

Substitute the value of density and volume of sodium chloride in the equation (2).

$M=(2.16 g/mL)(1 teaspoon)(5 mL1 teaspoon)=10.8 g$

The molality of a salt solution is given as,

$m=MsMmMW$

Where,

• $Ms$ represents the mass of the solute.

• $Mm$ represents the molar mass of the solute.

• $Mw$ represents the mass of the solvent.

Substitute the molar mass of sodium chloride, the mass of the sodium chloride and mass of the water in the above equation.

$m=10.8 g(58.44 g/mol)(3785.412 g)(1 kg1000 g)=(0.0488 mol/kg)(1 molal1 mol/kg )=0.0488 molal$

Therefore, the molality of the solution that is formed by dissolving one teaspoon of salt into four quarts of water is $0.0488 molal$ .

Substitute the value of molality of salt solution and $Kb$ in the equation (1).

$ΔTb=(0.0488 molal)(0.51°C/molal)=0.0249°C$

The elevation in the boiling point of water is very low. Therefore, the given amount of salt added to water does not substantially change the boiling point.

Answer 12

Conclusion

People who live at high altitudes are advised to add salt to water when boiling food like pasta to increase the boiling point of water The mole fraction of $NaCl$ is needed to raise the boiling point of $H2O$ by $3°C$ is $0.0957$ .The given amount of salt added to water does not substantially change the boiling point.

Answer 13

Ch. 7 - Prob. 7.1E Ch. 7 - Prob. 7.2E Ch. 7 - Prob. 7.3E Ch. 7 - Prob. 7.4E Ch. 7 - 7.5. For this chemical equilibrium in an enclosed...Ch. 7 - 7.6. The production of nitrogen gas for automobile...Ch. 7 - Prob. 7.7E Ch. 7 - Prob. 7.8E Ch. 7 - 7.9. For a solution of and in which , what are...Ch. 7 - Prob. 7.10E

Answer to Problem 7.71E

Explanation of Solution

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