CHEMISTRY:MOLECULAR NATURE...-ALEKS 360
CHEMISTRY:MOLECULAR NATURE...-ALEKS 360
8th Edition
ISBN: 9781259916083
Author: SILBERBERG
Publisher: MCG
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Chapter 13, Problem 13.120P
Interpretation Introduction

Interpretation:

The freezing point, boiling point and osmotic pressure of glucose are 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        (1)

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.

The boiling point of the substance is the temperature at which the vapor pressure of the liquid becomes equal to the atmospheric pressure and the liquid changes into a vapor. Liquids can change into vapors at temperatures below the boiling point through evaporation. It is the process that occurs on the liquid surface due to which it changes into vapors. Both boiling and freezing points are colligative properties because these depend on the number of moles of solute particles that are present in the substance.

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

ΔTb=ikbm        (2)

Here,

ΔTb is the change in boiling point.

i is van’t Hoff factor.

kb is the boiling point elevation constant.

m is the molality of the solution.

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:

=iMRT        (3)

Here,

is the osmotic pressure.

i is van’t Hoff factor.

M is the molarity of the solution.

R is universal gas constant.

T is the absolute temperature.

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

T(K)=T(°C)+273.15K

Expert Solution & Answer
Check Mark

Answer to Problem 13.120P

The freezing point, boiling point and osmotic pressure of glucose are 1.1 °C, 100.32 °C and 14 atm.

Explanation of Solution

Consider the mass of the solution to be 100g.

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

Mass of glucose=(Mass of solution)(mass %of glucosemass %of solution)        (4)

Substitute 100g for the mass of solution, 10 % for the mass % of glucose and 100 % for the mass % of the solution in equation (4).

Mass of glucose=(100g)(10 %100 %)=10g

The formula to calculate the moles of glucose is as follows:

Moles of glucose=Given mass of glucoseMolar mass of glucose        (5)

Substitute 10g for the given mass of glucose and 180.16g/mol for the molar mass of glucose in equation (5).

Moles of glucose=(10g)(1mol180.16g/mol)=0.055506217mol

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

Mass of glucose solution=Mass of glucose+Mass of water        (6)

Rearrange equation (6) to calculate the mass of water as follows:

Mass of water=Mass of glucose solutionMass of glucose        (7)

Substitute 100g for the mass of glucose solution and 10g for the mass of glucose in equation (7).

Mass of water=100 g10 g=90 g

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

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

Rearrange equation (8) to calculate the volume of the solution as follows:

Volume of solution=Mass of solutionDensity of solution        (9)

Substitute 100g for the mass of solution and 1.039g/mL for the density of the solution in equation (9).

Volume of solution=(100g)(1mL1.039g)(1L103mL)=0.096246L

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

Molarity=amount(mol)ofsolutevolume (L)ofsolution        (10)

Substitute 0.055506217mol for the amount of solute and 0.096246L for the volume of solution in equation (10) to calculate the molarity of glucose.

Molarity of glucose=0.055506217mol0.096246L=0.57671M

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

Molality=amount(mol)ofsolutemass(kg)ofsolvent        (11)

Substitute 0.055506217mol for the amount of solute and 90g for the mass of solvent in equation (11) to calculate the molality of glucose.

Molality of glucose=(0.055506217mol90g)(103g1kg)=0.6167357m

Glucose is a non-electrolyte so its van’t Hoff factor is 1.

Substitute 1 for i, 0.6167357m for m and 1.86 °C/m for kf in equation (1).

ΔTf=(1)(1.86 °C/m)(0.6167357m)=1.1471 °C

The formula to calculate the freezing point of glucose is as follows:

Freezingpointof glucose=Freezingpointof pure H2OΔTf        (12)

Substitute 0 °C for the freezing point of pure water and 1.1471 °C for ΔTf in equation (12).

Freezingpointof glucose=0 °C1.1471 °C=1.1471 °C=1.1 °C

Substitute 1 for i, 0.6167357m for m and 0.512 °C/m for kb in equation (2).

ΔTb=(1)(0.512 °C/m)(0.6167357m)=0.3157687 °C

The formula to calculate the boiling point of glucose is as follows:

Boilingpointof glucose=Boilingpointof pure H2O+ΔTb        (13)

Substitute 100 °C for the boiling point of pure water and 0.3157687 °C for ΔTb in equation (12).

Boilingpointof glucose=100 °C+0.3157687 °C=100.3157687 °C=100.32 °C

Substitute 0.57671M for M, 1 for i, 0.0821Latm/molK for R and 20 °C for T in equation (3).

=(1)(0.57671M)(0.0821Latm/molK)(20 °C+273.15)K=13.8729atm=14 atm

Conclusion

The freezing point, boiling point and osmotic pressure of glucose are 1.1 °C, 100.32 °C and 14 atm.

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Chapter 13 Solutions

CHEMISTRY:MOLECULAR NATURE...-ALEKS 360

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Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY