The molar mass of the fragment is to be calculated. Concept introduction: The osmotic pressure is defined as the measure of the tendency of a solution to take in pure solvent via osmosis. It is defined as the minimum pressure that is to be applied to the solution to prevent the inward flow of the pure solvent across the semipermeable membrane. Osmosis occurs when two solutions have different concentrations of solute and are separated by a semipermeable membrane. The formula to calculate the osmotic pressure of the solution is as follows: ∏ = M R T (1) Here, ∏ is the osmotic pressure. M is the molarity of the solution. R is universal gas constant. T is the absolute temperature. Molarity is defined as the number of moles of solute that are dissolved in one litre of solution. It is represented by M and its unit is mol/L . The formula to calculate the molarity of the solution is as follows: Molarity = amount ( mol ) of solute volume ( L ) of solution (2) The conversion factor to convert ° C to ° K is as follows: T ( ° K ) = T ( ° C ) + 273.15 K

Question

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Answer 1

Question

Chapter 13, Problem 13.139P

(a)

Interpretation Introduction

Interpretation:

The molar mass of the fragment is to be calculated.

Concept introduction:

The osmotic pressure is defined as the measure of the tendency of a solution to take in pure solvent via osmosis. It is defined as the minimum pressure that is to be applied to the solution to prevent the inward flow of the pure solvent across the semipermeable membrane. Osmosis occurs when two solutions have different concentrations of solute and are separated by a semipermeable membrane.

The formula to calculate the osmotic pressure of the solution is as follows:

∏=MRT (1)

Here, ∏ is the osmotic pressure.

M is the molarity of the solution.

R is universal gas constant.

T is the absolute temperature.

Molarity is defined as the number of moles of solute that are dissolved in one litre of solution. It is represented by M and its unit is mol/L.

The formula to calculate the molarity of the solution is as follows:

Molarity=amount (mol) of solute volume (L) of solution (2)

The conversion factor to convert °C to °K is as follows:

T(°K)=T(°C)+273.15 K

(a)

Expert Solution

Answer to Problem 13.139P

The molar mass of the fragment is 1.82×104 g/mol.

Explanation of Solution

Rearrange equation (1) to calculate the molarity of the solution as follows:

M=∏RT (3)

Substitute 0.340 torr for ∏, 0.0821 L⋅atm/mol⋅K for R and 25°C for T in equation (3).

M=(0.340 torr)(1 atm760 torr)(0.0821 L⋅atm/mol⋅K)(25°C+273.15) K=1.828544×10−5 M

Rearrange equation (2) to calculate the amount of solute as follows:

Amount of solute=(Molarity)(Volume of solution) (4)

Substitute 1.828544×10−5 M for the molarity and 30 mL for the volume of solution in equation (4).

Amount of solute=(1.828544×10−5 mol1 L)(30 mL)(10−3 L1 mL)=5.48563×10−5mol

The molar mass of the compound is calculated as follows:

Molar mass=Given massNumber of moles (5)

Substitute 5.48563×10−5mol for the number of moles and 10 mg for the given mass in equation (5).

Molar mass=(10 mg)(10−3g1 mg)5.48563×10−5mol=1.82294×104 g/mol=1.82×104 g/mol

Conclusion

The molar mass of the fragment is 1.82×104 g/mol.

(b)

Interpretation Introduction

Interpretation:

The depression in freezing point is to be calculated.

Concept introduction:

The freezing point is the temperature at which both the solid and liquid phases coexist in equilibrium. It is the temperature at which the vapor pressure of the substance in the liquid state becomes equal to the vapor pressure in a solid state.

The formula to calculate the change in freezing point is as follows:

ΔTf=ikfm (6)

Here, ΔTf is the change in freezing point.

i is van’t Hoff factor.

kf is the freezing point depression constant.

m is the molality of the solution.

Molality is the measure of the concentration of solute in the solution. It is the amount of solute that is dissolved in one kilogram of the solvent. It is represented by m and its unit is moles per kilograms. The solute is the substance that is present in a smaller amount and solvent is the substance that is present in a larger amount.

The formula to calculate the density of the solution is as follows:

Density of solution=Mass of solutionVolume of solution (7)

(b)

Expert Solution

Answer to Problem 13.139P

The freezing point of the solution is −3.41×10−5°C.

Explanation of Solution

Rearrange equation (7) to calculate the mass of the solution as follows:

Mass of solution=(Density of solution)(Volume of solution) (8)

Substitute 0.997 g/mL for the density of the solution and 30 mL for the volume of solution in equation (8).

Mass of solution=(0.997 g1 mL)(30 mL)=29.91 g

The formula to calculate the mass of the solution is as follows:

Mass of solution=Mass of solute+Mass of solvent (9)

Rearrange equation (9) to calculate the mass of the solvent as follows:

Mass of solvent=Mass of solution−Mass of solute (10)

Substitute 29.91 g for the mass of solution and 10 mg for the mass of solute in equation (10) to calculate the mass of water.

Mass of water=(29.91 g−(10 mg)(10−3 g1 mg))(1 kg103g)=0.0299 kg

The formula to calculate the molality of the solution is as follows:

Molality=amount (mol) of solute mass (kg) of solvent (11)

Substitute 5.48563×10−5mol for the number of moles and 0.0299 kg for the mass of solvent in equation (11).

Molality=5.48563×10−5mol 0.0299 kg=1.83466×10−5 m

Substitute 1.83466×10−5 m for m, 1 for i, 1.86°C/m for kf in equation (6).

ΔTf=ikfm=(1)(1.86°C/m)(1.83466×10−5 m)=3.412×10−5°C=3.41×10−5°C

But this is the change in freezing point of the solution.

The freezing point of the solution is calculated as follows:

Tf=Tf(solvent)−Tf(solution) (12)

Substitute 0°C for Tf(solvent) and 3.412×10−5°C for Tf(solution) in equation (120.

Tf=Tf(solvent)−Tf(solution)=0°C−3.412×10−5°C=−3.412×10−5°C=−3.41×10−5°C

Conclusion

The freezing point of the solution is −3.41×10−5°C.

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Answer 2

Question

Chapter 13, Problem 13.139P

(a)

Interpretation Introduction

Interpretation:

The molar mass of the fragment is to be calculated.

Concept introduction:

The osmotic pressure is defined as the measure of the tendency of a solution to take in pure solvent via osmosis. It is defined as the minimum pressure that is to be applied to the solution to prevent the inward flow of the pure solvent across the semipermeable membrane. Osmosis occurs when two solutions have different concentrations of solute and are separated by a semipermeable membrane.

The formula to calculate the osmotic pressure of the solution is as follows:

∏=MRT (1)

Here, ∏ is the osmotic pressure.

M is the molarity of the solution.

R is universal gas constant.

T is the absolute temperature.

Molarity is defined as the number of moles of solute that are dissolved in one litre of solution. It is represented by M and its unit is mol/L.

The formula to calculate the molarity of the solution is as follows:

Molarity=amount (mol) of solute volume (L) of solution (2)

The conversion factor to convert °C to °K is as follows:

T(°K)=T(°C)+273.15 K

(a)

Expert Solution

Answer to Problem 13.139P

The molar mass of the fragment is 1.82×104 g/mol.

Explanation of Solution

Rearrange equation (1) to calculate the molarity of the solution as follows:

M=∏RT (3)

Substitute 0.340 torr for ∏, 0.0821 L⋅atm/mol⋅K for R and 25°C for T in equation (3).

M=(0.340 torr)(1 atm760 torr)(0.0821 L⋅atm/mol⋅K)(25°C+273.15) K=1.828544×10−5 M

Rearrange equation (2) to calculate the amount of solute as follows:

Amount of solute=(Molarity)(Volume of solution) (4)

Substitute 1.828544×10−5 M for the molarity and 30 mL for the volume of solution in equation (4).

Amount of solute=(1.828544×10−5 mol1 L)(30 mL)(10−3 L1 mL)=5.48563×10−5mol

The molar mass of the compound is calculated as follows:

Molar mass=Given massNumber of moles (5)

Substitute 5.48563×10−5mol for the number of moles and 10 mg for the given mass in equation (5).

Molar mass=(10 mg)(10−3g1 mg)5.48563×10−5mol=1.82294×104 g/mol=1.82×104 g/mol

Conclusion

The molar mass of the fragment is 1.82×104 g/mol.

(b)

Interpretation Introduction

Interpretation:

The depression in freezing point is to be calculated.

Concept introduction:

The freezing point is the temperature at which both the solid and liquid phases coexist in equilibrium. It is the temperature at which the vapor pressure of the substance in the liquid state becomes equal to the vapor pressure in a solid state.

The formula to calculate the change in freezing point is as follows:

ΔTf=ikfm (6)

Here, ΔTf is the change in freezing point.

i is van’t Hoff factor.

kf is the freezing point depression constant.

m is the molality of the solution.

Molality is the measure of the concentration of solute in the solution. It is the amount of solute that is dissolved in one kilogram of the solvent. It is represented by m and its unit is moles per kilograms. The solute is the substance that is present in a smaller amount and solvent is the substance that is present in a larger amount.

The formula to calculate the density of the solution is as follows:

Density of solution=Mass of solutionVolume of solution (7)

(b)

Expert Solution

Answer to Problem 13.139P

The freezing point of the solution is −3.41×10−5°C.

Explanation of Solution

Rearrange equation (7) to calculate the mass of the solution as follows:

Mass of solution=(Density of solution)(Volume of solution) (8)

Substitute 0.997 g/mL for the density of the solution and 30 mL for the volume of solution in equation (8).

Mass of solution=(0.997 g1 mL)(30 mL)=29.91 g

The formula to calculate the mass of the solution is as follows:

Mass of solution=Mass of solute+Mass of solvent (9)

Rearrange equation (9) to calculate the mass of the solvent as follows:

Mass of solvent=Mass of solution−Mass of solute (10)

Substitute 29.91 g for the mass of solution and 10 mg for the mass of solute in equation (10) to calculate the mass of water.

Mass of water=(29.91 g−(10 mg)(10−3 g1 mg))(1 kg103g)=0.0299 kg

The formula to calculate the molality of the solution is as follows:

Molality=amount (mol) of solute mass (kg) of solvent (11)

Substitute 5.48563×10−5mol for the number of moles and 0.0299 kg for the mass of solvent in equation (11).

Molality=5.48563×10−5mol 0.0299 kg=1.83466×10−5 m

Substitute 1.83466×10−5 m for m, 1 for i, 1.86°C/m for kf in equation (6).

ΔTf=ikfm=(1)(1.86°C/m)(1.83466×10−5 m)=3.412×10−5°C=3.41×10−5°C

But this is the change in freezing point of the solution.

The freezing point of the solution is calculated as follows:

Tf=Tf(solvent)−Tf(solution) (12)

Substitute 0°C for Tf(solvent) and 3.412×10−5°C for Tf(solution) in equation (120.

Tf=Tf(solvent)−Tf(solution)=0°C−3.412×10−5°C=−3.412×10−5°C=−3.41×10−5°C

Conclusion

The freezing point of the solution is −3.41×10−5°C.

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Answer 3

Question

Chapter 13, Problem 13.139P

(a)

Interpretation Introduction

Interpretation:

The molar mass of the fragment is to be calculated.

Concept introduction:

The osmotic pressure is defined as the measure of the tendency of a solution to take in pure solvent via osmosis. It is defined as the minimum pressure that is to be applied to the solution to prevent the inward flow of the pure solvent across the semipermeable membrane. Osmosis occurs when two solutions have different concentrations of solute and are separated by a semipermeable membrane.

The formula to calculate the osmotic pressure of the solution is as follows:

∏=MRT (1)

Here, ∏ is the osmotic pressure.

M is the molarity of the solution.

R is universal gas constant.

T is the absolute temperature.

Molarity is defined as the number of moles of solute that are dissolved in one litre of solution. It is represented by M and its unit is mol/L.

The formula to calculate the molarity of the solution is as follows:

Molarity=amount (mol) of solute volume (L) of solution (2)

The conversion factor to convert °C to °K is as follows:

T(°K)=T(°C)+273.15 K

(a)

Expert Solution

Answer to Problem 13.139P

The molar mass of the fragment is 1.82×104 g/mol.

Explanation of Solution

Rearrange equation (1) to calculate the molarity of the solution as follows:

M=∏RT (3)

Substitute 0.340 torr for ∏, 0.0821 L⋅atm/mol⋅K for R and 25°C for T in equation (3).

M=(0.340 torr)(1 atm760 torr)(0.0821 L⋅atm/mol⋅K)(25°C+273.15) K=1.828544×10−5 M

Rearrange equation (2) to calculate the amount of solute as follows:

Amount of solute=(Molarity)(Volume of solution) (4)

Substitute 1.828544×10−5 M for the molarity and 30 mL for the volume of solution in equation (4).

Amount of solute=(1.828544×10−5 mol1 L)(30 mL)(10−3 L1 mL)=5.48563×10−5mol

The molar mass of the compound is calculated as follows:

Molar mass=Given massNumber of moles (5)

Substitute 5.48563×10−5mol for the number of moles and 10 mg for the given mass in equation (5).

Molar mass=(10 mg)(10−3g1 mg)5.48563×10−5mol=1.82294×104 g/mol=1.82×104 g/mol

Conclusion

The molar mass of the fragment is 1.82×104 g/mol.

(b)

Interpretation Introduction

Interpretation:

The depression in freezing point is to be calculated.

Concept introduction:

The freezing point is the temperature at which both the solid and liquid phases coexist in equilibrium. It is the temperature at which the vapor pressure of the substance in the liquid state becomes equal to the vapor pressure in a solid state.

The formula to calculate the change in freezing point is as follows:

ΔTf=ikfm (6)

Here, ΔTf is the change in freezing point.

i is van’t Hoff factor.

kf is the freezing point depression constant.

m is the molality of the solution.

Molality is the measure of the concentration of solute in the solution. It is the amount of solute that is dissolved in one kilogram of the solvent. It is represented by m and its unit is moles per kilograms. The solute is the substance that is present in a smaller amount and solvent is the substance that is present in a larger amount.

The formula to calculate the density of the solution is as follows:

Density of solution=Mass of solutionVolume of solution (7)

(b)

Expert Solution

Answer to Problem 13.139P

The freezing point of the solution is −3.41×10−5°C.

Explanation of Solution

Rearrange equation (7) to calculate the mass of the solution as follows:

Mass of solution=(Density of solution)(Volume of solution) (8)

Substitute 0.997 g/mL for the density of the solution and 30 mL for the volume of solution in equation (8).

Mass of solution=(0.997 g1 mL)(30 mL)=29.91 g

The formula to calculate the mass of the solution is as follows:

Mass of solution=Mass of solute+Mass of solvent (9)

Rearrange equation (9) to calculate the mass of the solvent as follows:

Mass of solvent=Mass of solution−Mass of solute (10)

Substitute 29.91 g for the mass of solution and 10 mg for the mass of solute in equation (10) to calculate the mass of water.

Mass of water=(29.91 g−(10 mg)(10−3 g1 mg))(1 kg103g)=0.0299 kg

The formula to calculate the molality of the solution is as follows:

Molality=amount (mol) of solute mass (kg) of solvent (11)

Substitute 5.48563×10−5mol for the number of moles and 0.0299 kg for the mass of solvent in equation (11).

Molality=5.48563×10−5mol 0.0299 kg=1.83466×10−5 m

Substitute 1.83466×10−5 m for m, 1 for i, 1.86°C/m for kf in equation (6).

ΔTf=ikfm=(1)(1.86°C/m)(1.83466×10−5 m)=3.412×10−5°C=3.41×10−5°C

But this is the change in freezing point of the solution.

The freezing point of the solution is calculated as follows:

Tf=Tf(solvent)−Tf(solution) (12)

Substitute 0°C for Tf(solvent) and 3.412×10−5°C for Tf(solution) in equation (120.

Tf=Tf(solvent)−Tf(solution)=0°C−3.412×10−5°C=−3.412×10−5°C=−3.41×10−5°C

Conclusion

The freezing point of the solution is −3.41×10−5°C.

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Answer 4

Question

Chapter 13, Problem 13.139P

(a)

Interpretation Introduction

Interpretation:

The molar mass of the fragment is to be calculated.

Concept introduction:

The osmotic pressure is defined as the measure of the tendency of a solution to take in pure solvent via osmosis. It is defined as the minimum pressure that is to be applied to the solution to prevent the inward flow of the pure solvent across the semipermeable membrane. Osmosis occurs when two solutions have different concentrations of solute and are separated by a semipermeable membrane.

The formula to calculate the osmotic pressure of the solution is as follows:

∏=MRT (1)

Here, ∏ is the osmotic pressure.

M is the molarity of the solution.

R is universal gas constant.

T is the absolute temperature.

Molarity is defined as the number of moles of solute that are dissolved in one litre of solution. It is represented by M and its unit is mol/L.

The formula to calculate the molarity of the solution is as follows:

Molarity=amount (mol) of solute volume (L) of solution (2)

The conversion factor to convert °C to °K is as follows:

T(°K)=T(°C)+273.15 K

(a)

Expert Solution

Answer to Problem 13.139P

The molar mass of the fragment is 1.82×104 g/mol.

Explanation of Solution

Rearrange equation (1) to calculate the molarity of the solution as follows:

M=∏RT (3)

Substitute 0.340 torr for ∏, 0.0821 L⋅atm/mol⋅K for R and 25°C for T in equation (3).

M=(0.340 torr)(1 atm760 torr)(0.0821 L⋅atm/mol⋅K)(25°C+273.15) K=1.828544×10−5 M

Rearrange equation (2) to calculate the amount of solute as follows:

Amount of solute=(Molarity)(Volume of solution) (4)

Substitute 1.828544×10−5 M for the molarity and 30 mL for the volume of solution in equation (4).

Amount of solute=(1.828544×10−5 mol1 L)(30 mL)(10−3 L1 mL)=5.48563×10−5mol

The molar mass of the compound is calculated as follows:

Molar mass=Given massNumber of moles (5)

Substitute 5.48563×10−5mol for the number of moles and 10 mg for the given mass in equation (5).

Molar mass=(10 mg)(10−3g1 mg)5.48563×10−5mol=1.82294×104 g/mol=1.82×104 g/mol

Conclusion

The molar mass of the fragment is 1.82×104 g/mol.

(b)

Interpretation Introduction

Interpretation:

The depression in freezing point is to be calculated.

Concept introduction:

The freezing point is the temperature at which both the solid and liquid phases coexist in equilibrium. It is the temperature at which the vapor pressure of the substance in the liquid state becomes equal to the vapor pressure in a solid state.

The formula to calculate the change in freezing point is as follows:

ΔTf=ikfm (6)

Here, ΔTf is the change in freezing point.

i is van’t Hoff factor.

kf is the freezing point depression constant.

m is the molality of the solution.

Molality is the measure of the concentration of solute in the solution. It is the amount of solute that is dissolved in one kilogram of the solvent. It is represented by m and its unit is moles per kilograms. The solute is the substance that is present in a smaller amount and solvent is the substance that is present in a larger amount.

The formula to calculate the density of the solution is as follows:

Density of solution=Mass of solutionVolume of solution (7)

(b)

Expert Solution

Answer to Problem 13.139P

The freezing point of the solution is −3.41×10−5°C.

Explanation of Solution

Rearrange equation (7) to calculate the mass of the solution as follows:

Mass of solution=(Density of solution)(Volume of solution) (8)

Substitute 0.997 g/mL for the density of the solution and 30 mL for the volume of solution in equation (8).

Mass of solution=(0.997 g1 mL)(30 mL)=29.91 g

The formula to calculate the mass of the solution is as follows:

Mass of solution=Mass of solute+Mass of solvent (9)

Rearrange equation (9) to calculate the mass of the solvent as follows:

Mass of solvent=Mass of solution−Mass of solute (10)

Substitute 29.91 g for the mass of solution and 10 mg for the mass of solute in equation (10) to calculate the mass of water.

Mass of water=(29.91 g−(10 mg)(10−3 g1 mg))(1 kg103g)=0.0299 kg

The formula to calculate the molality of the solution is as follows:

Molality=amount (mol) of solute mass (kg) of solvent (11)

Substitute 5.48563×10−5mol for the number of moles and 0.0299 kg for the mass of solvent in equation (11).

Molality=5.48563×10−5mol 0.0299 kg=1.83466×10−5 m

Substitute 1.83466×10−5 m for m, 1 for i, 1.86°C/m for kf in equation (6).

ΔTf=ikfm=(1)(1.86°C/m)(1.83466×10−5 m)=3.412×10−5°C=3.41×10−5°C

But this is the change in freezing point of the solution.

The freezing point of the solution is calculated as follows:

Tf=Tf(solvent)−Tf(solution) (12)

Substitute 0°C for Tf(solvent) and 3.412×10−5°C for Tf(solution) in equation (120.

Tf=Tf(solvent)−Tf(solution)=0°C−3.412×10−5°C=−3.412×10−5°C=−3.41×10−5°C

Conclusion

The freezing point of the solution is −3.41×10−5°C.

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Answer 5

Chapter 13, Problem 13.139P

(a)

Interpretation Introduction

Interpretation:

The molar mass of the fragment is to be calculated.

Concept introduction:

The osmotic pressure is defined as the measure of the tendency of a solution to take in pure solvent via osmosis. It is defined as the minimum pressure that is to be applied to the solution to prevent the inward flow of the pure solvent across the semipermeable membrane. Osmosis occurs when two solutions have different concentrations of solute and are separated by a semipermeable membrane.

The formula to calculate the osmotic pressure of the solution is as follows:

∏=MRT (1)

Here, ∏ is the osmotic pressure.

M is the molarity of the solution.

R is universal gas constant.

T is the absolute temperature.

Molarity is defined as the number of moles of solute that are dissolved in one litre of solution. It is represented by M and its unit is mol/L.

The formula to calculate the molarity of the solution is as follows:

Molarity=amount (mol) of solute volume (L) of solution (2)

The conversion factor to convert °C to °K is as follows:

T(°K)=T(°C)+273.15 K

(a)

Expert Solution

Answer to Problem 13.139P

The molar mass of the fragment is 1.82×104 g/mol.

Explanation of Solution

Rearrange equation (1) to calculate the molarity of the solution as follows:

M=∏RT (3)

Substitute 0.340 torr for ∏, 0.0821 L⋅atm/mol⋅K for R and 25°C for T in equation (3).

M=(0.340 torr)(1 atm760 torr)(0.0821 L⋅atm/mol⋅K)(25°C+273.15) K=1.828544×10−5 M

Rearrange equation (2) to calculate the amount of solute as follows:

Amount of solute=(Molarity)(Volume of solution) (4)

Substitute 1.828544×10−5 M for the molarity and 30 mL for the volume of solution in equation (4).

Amount of solute=(1.828544×10−5 mol1 L)(30 mL)(10−3 L1 mL)=5.48563×10−5mol

The molar mass of the compound is calculated as follows:

Molar mass=Given massNumber of moles (5)

Substitute 5.48563×10−5mol for the number of moles and 10 mg for the given mass in equation (5).

Molar mass=(10 mg)(10−3g1 mg)5.48563×10−5mol=1.82294×104 g/mol=1.82×104 g/mol

Conclusion

The molar mass of the fragment is 1.82×104 g/mol.

(b)

Interpretation Introduction

Interpretation:

The depression in freezing point is to be calculated.

Concept introduction:

The freezing point is the temperature at which both the solid and liquid phases coexist in equilibrium. It is the temperature at which the vapor pressure of the substance in the liquid state becomes equal to the vapor pressure in a solid state.

The formula to calculate the change in freezing point is as follows:

ΔTf=ikfm (6)

Here, ΔTf is the change in freezing point.

i is van’t Hoff factor.

kf is the freezing point depression constant.

m is the molality of the solution.

Molality is the measure of the concentration of solute in the solution. It is the amount of solute that is dissolved in one kilogram of the solvent. It is represented by m and its unit is moles per kilograms. The solute is the substance that is present in a smaller amount and solvent is the substance that is present in a larger amount.

The formula to calculate the density of the solution is as follows:

Density of solution=Mass of solutionVolume of solution (7)

(b)

Expert Solution

Answer to Problem 13.139P

The freezing point of the solution is −3.41×10−5°C.

Explanation of Solution

Rearrange equation (7) to calculate the mass of the solution as follows:

Mass of solution=(Density of solution)(Volume of solution) (8)

Substitute 0.997 g/mL for the density of the solution and 30 mL for the volume of solution in equation (8).

Mass of solution=(0.997 g1 mL)(30 mL)=29.91 g

The formula to calculate the mass of the solution is as follows:

Mass of solution=Mass of solute+Mass of solvent (9)

Rearrange equation (9) to calculate the mass of the solvent as follows:

Mass of solvent=Mass of solution−Mass of solute (10)

Substitute 29.91 g for the mass of solution and 10 mg for the mass of solute in equation (10) to calculate the mass of water.

Mass of water=(29.91 g−(10 mg)(10−3 g1 mg))(1 kg103g)=0.0299 kg

The formula to calculate the molality of the solution is as follows:

Molality=amount (mol) of solute mass (kg) of solvent (11)

Substitute 5.48563×10−5mol for the number of moles and 0.0299 kg for the mass of solvent in equation (11).

Molality=5.48563×10−5mol 0.0299 kg=1.83466×10−5 m

Substitute 1.83466×10−5 m for m, 1 for i, 1.86°C/m for kf in equation (6).

ΔTf=ikfm=(1)(1.86°C/m)(1.83466×10−5 m)=3.412×10−5°C=3.41×10−5°C

But this is the change in freezing point of the solution.

The freezing point of the solution is calculated as follows:

Tf=Tf(solvent)−Tf(solution) (12)

Substitute 0°C for Tf(solvent) and 3.412×10−5°C for Tf(solution) in equation (120.

Tf=Tf(solvent)−Tf(solution)=0°C−3.412×10−5°C=−3.412×10−5°C=−3.41×10−5°C

Conclusion

The freezing point of the solution is −3.41×10−5°C.

Answer 6

Chapter 13, Problem 13.139P

(a)

Interpretation Introduction

Interpretation:

The molar mass of the fragment is to be calculated.

Concept introduction:

The osmotic pressure is defined as the measure of the tendency of a solution to take in pure solvent via osmosis. It is defined as the minimum pressure that is to be applied to the solution to prevent the inward flow of the pure solvent across the semipermeable membrane. Osmosis occurs when two solutions have different concentrations of solute and are separated by a semipermeable membrane.

The formula to calculate the osmotic pressure of the solution is as follows:

∏=MRT (1)

Here, ∏ is the osmotic pressure.

M is the molarity of the solution.

R is universal gas constant.

T is the absolute temperature.

Molarity is defined as the number of moles of solute that are dissolved in one litre of solution. It is represented by M and its unit is mol/L.

The formula to calculate the molarity of the solution is as follows:

Molarity=amount (mol) of solute volume (L) of solution (2)

The conversion factor to convert °C to °K is as follows:

T(°K)=T(°C)+273.15 K

(a)

Expert Solution

Answer to Problem 13.139P

The molar mass of the fragment is 1.82×104 g/mol.

Explanation of Solution

Rearrange equation (1) to calculate the molarity of the solution as follows:

M=∏RT (3)

Substitute 0.340 torr for ∏, 0.0821 L⋅atm/mol⋅K for R and 25°C for T in equation (3).

M=(0.340 torr)(1 atm760 torr)(0.0821 L⋅atm/mol⋅K)(25°C+273.15) K=1.828544×10−5 M

Rearrange equation (2) to calculate the amount of solute as follows:

Amount of solute=(Molarity)(Volume of solution) (4)

Substitute 1.828544×10−5 M for the molarity and 30 mL for the volume of solution in equation (4).

Amount of solute=(1.828544×10−5 mol1 L)(30 mL)(10−3 L1 mL)=5.48563×10−5mol

The molar mass of the compound is calculated as follows:

Molar mass=Given massNumber of moles (5)

Substitute 5.48563×10−5mol for the number of moles and 10 mg for the given mass in equation (5).

Molar mass=(10 mg)(10−3g1 mg)5.48563×10−5mol=1.82294×104 g/mol=1.82×104 g/mol

Conclusion

The molar mass of the fragment is 1.82×104 g/mol.

Answer 7

Chapter 13, Problem 13.139P

(a)

Interpretation Introduction

Interpretation:

The molar mass of the fragment is to be calculated.

Concept introduction:

The osmotic pressure is defined as the measure of the tendency of a solution to take in pure solvent via osmosis. It is defined as the minimum pressure that is to be applied to the solution to prevent the inward flow of the pure solvent across the semipermeable membrane. Osmosis occurs when two solutions have different concentrations of solute and are separated by a semipermeable membrane.

The formula to calculate the osmotic pressure of the solution is as follows:

∏=MRT (1)

Here, ∏ is the osmotic pressure.

M is the molarity of the solution.

R is universal gas constant.

T is the absolute temperature.

Molarity is defined as the number of moles of solute that are dissolved in one litre of solution. It is represented by M and its unit is mol/L.

The formula to calculate the molarity of the solution is as follows:

Molarity=amount (mol) of solute volume (L) of solution (2)

The conversion factor to convert °C to °K is as follows:

T(°K)=T(°C)+273.15 K

Answer 8

(a)

Expert Solution

Answer to Problem 13.139P

The molar mass of the fragment is 1.82×104 g/mol.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Rearrange equation (1) to calculate the molarity of the solution as follows:

M=∏RT (3)

Substitute 0.340 torr for ∏, 0.0821 L⋅atm/mol⋅K for R and 25°C for T in equation (3).

M=(0.340 torr)(1 atm760 torr)(0.0821 L⋅atm/mol⋅K)(25°C+273.15) K=1.828544×10−5 M

Rearrange equation (2) to calculate the amount of solute as follows:

Amount of solute=(Molarity)(Volume of solution) (4)

Substitute 1.828544×10−5 M for the molarity and 30 mL for the volume of solution in equation (4).

Amount of solute=(1.828544×10−5 mol1 L)(30 mL)(10−3 L1 mL)=5.48563×10−5mol

The molar mass of the compound is calculated as follows:

Molar mass=Given massNumber of moles (5)

Substitute 5.48563×10−5mol for the number of moles and 10 mg for the given mass in equation (5).

Molar mass=(10 mg)(10−3g1 mg)5.48563×10−5mol=1.82294×104 g/mol=1.82×104 g/mol

Answer 13

Conclusion

The molar mass of the fragment is 1.82×104 g/mol.

Answer 14

(b)

Interpretation Introduction

Interpretation:

The depression in freezing point is to be calculated.

Concept introduction:

The freezing point is the temperature at which both the solid and liquid phases coexist in equilibrium. It is the temperature at which the vapor pressure of the substance in the liquid state becomes equal to the vapor pressure in a solid state.

The formula to calculate the change in freezing point is as follows:

ΔTf=ikfm (6)

Here, ΔTf is the change in freezing point.

i is van’t Hoff factor.

kf is the freezing point depression constant.

m is the molality of the solution.

Molality is the measure of the concentration of solute in the solution. It is the amount of solute that is dissolved in one kilogram of the solvent. It is represented by m and its unit is moles per kilograms. The solute is the substance that is present in a smaller amount and solvent is the substance that is present in a larger amount.

The formula to calculate the density of the solution is as follows:

Density of solution=Mass of solutionVolume of solution (7)

(b)

Expert Solution

Answer to Problem 13.139P

The freezing point of the solution is −3.41×10−5°C.

Explanation of Solution

Rearrange equation (7) to calculate the mass of the solution as follows:

Mass of solution=(Density of solution)(Volume of solution) (8)

Substitute 0.997 g/mL for the density of the solution and 30 mL for the volume of solution in equation (8).

Mass of solution=(0.997 g1 mL)(30 mL)=29.91 g

The formula to calculate the mass of the solution is as follows:

Mass of solution=Mass of solute+Mass of solvent (9)

Rearrange equation (9) to calculate the mass of the solvent as follows:

Mass of solvent=Mass of solution−Mass of solute (10)

Substitute 29.91 g for the mass of solution and 10 mg for the mass of solute in equation (10) to calculate the mass of water.

Mass of water=(29.91 g−(10 mg)(10−3 g1 mg))(1 kg103g)=0.0299 kg

The formula to calculate the molality of the solution is as follows:

Molality=amount (mol) of solute mass (kg) of solvent (11)

Substitute 5.48563×10−5mol for the number of moles and 0.0299 kg for the mass of solvent in equation (11).

Molality=5.48563×10−5mol 0.0299 kg=1.83466×10−5 m

Substitute 1.83466×10−5 m for m, 1 for i, 1.86°C/m for kf in equation (6).

ΔTf=ikfm=(1)(1.86°C/m)(1.83466×10−5 m)=3.412×10−5°C=3.41×10−5°C

But this is the change in freezing point of the solution.

The freezing point of the solution is calculated as follows:

Tf=Tf(solvent)−Tf(solution) (12)

Substitute 0°C for Tf(solvent) and 3.412×10−5°C for Tf(solution) in equation (120.

Tf=Tf(solvent)−Tf(solution)=0°C−3.412×10−5°C=−3.412×10−5°C=−3.41×10−5°C

Conclusion

The freezing point of the solution is −3.41×10−5°C.

Answer 15

(b)

Interpretation Introduction

Interpretation:

The depression in freezing point is to be calculated.

Concept introduction:

The freezing point is the temperature at which both the solid and liquid phases coexist in equilibrium. It is the temperature at which the vapor pressure of the substance in the liquid state becomes equal to the vapor pressure in a solid state.

The formula to calculate the change in freezing point is as follows:

ΔTf=ikfm (6)

Here, ΔTf is the change in freezing point.

i is van’t Hoff factor.

kf is the freezing point depression constant.

m is the molality of the solution.

Molality is the measure of the concentration of solute in the solution. It is the amount of solute that is dissolved in one kilogram of the solvent. It is represented by m and its unit is moles per kilograms. The solute is the substance that is present in a smaller amount and solvent is the substance that is present in a larger amount.

The formula to calculate the density of the solution is as follows:

Density of solution=Mass of solutionVolume of solution (7)

Answer 16

(b)

Expert Solution

Answer to Problem 13.139P

The freezing point of the solution is −3.41×10−5°C.

Answer 17

(b)

Expert Solution

Answer 18

(b)

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

Rearrange equation (7) to calculate the mass of the solution as follows:

Mass of solution=(Density of solution)(Volume of solution) (8)

Substitute 0.997 g/mL for the density of the solution and 30 mL for the volume of solution in equation (8).

Mass of solution=(0.997 g1 mL)(30 mL)=29.91 g

The formula to calculate the mass of the solution is as follows:

Mass of solution=Mass of solute+Mass of solvent (9)

Rearrange equation (9) to calculate the mass of the solvent as follows:

Mass of solvent=Mass of solution−Mass of solute (10)

Substitute 29.91 g for the mass of solution and 10 mg for the mass of solute in equation (10) to calculate the mass of water.

Mass of water=(29.91 g−(10 mg)(10−3 g1 mg))(1 kg103g)=0.0299 kg

The formula to calculate the molality of the solution is as follows:

Molality=amount (mol) of solute mass (kg) of solvent (11)

Substitute 5.48563×10−5mol for the number of moles and 0.0299 kg for the mass of solvent in equation (11).

Molality=5.48563×10−5mol 0.0299 kg=1.83466×10−5 m

Substitute 1.83466×10−5 m for m, 1 for i, 1.86°C/m for kf in equation (6).

ΔTf=ikfm=(1)(1.86°C/m)(1.83466×10−5 m)=3.412×10−5°C=3.41×10−5°C

But this is the change in freezing point of the solution.

The freezing point of the solution is calculated as follows:

Tf=Tf(solvent)−Tf(solution) (12)

Substitute 0°C for Tf(solvent) and 3.412×10−5°C for Tf(solution) in equation (120.

Tf=Tf(solvent)−Tf(solution)=0°C−3.412×10−5°C=−3.412×10−5°C=−3.41×10−5°C

Answer 21

Conclusion

The freezing point of the solution is −3.41×10−5°C.

Answer 22

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