   # An aqueous solution containing 10.0 g of starch per liter has an osmotic pressure of 3.8 mm Hg at 25 °C. (a) What is the average molar mass of starch? (The result is an average because not all starch molecules are identical.) (b) What is the freezing point of the solution? Would it be easy to determine the molecular weight of starch by measuring the freezing point depression? (Assume that the molarity and molality are the same for this solution.) ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 13, Problem 71GQ
Textbook Problem
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## An aqueous solution containing 10.0 g of starch per liter has an osmotic pressure of 3.8 mm Hg at 25 °C. (a) What is the average molar mass of starch? (The result is an average because not all starch molecules are identical.) (b) What is the freezing point of the solution? Would it be easy to determine the molecular weight of starch by measuring the freezing point depression? (Assume that the molarity and molality are the same for this solution.)

(a)

Interpretation Introduction

Interpretation: The average molar mass of starch of the solution has to be determined.

Concept introduction:

Colligative properties: Properties of solutions which having influence on the concentration of the solute in it. Colligative properties are,

• Decrease in the vapor pressure
• Increase in the boiling point
• Decline in the freezing point
• Osmotic pressure

Osmotic pressure: The pressure created by the column of solution for the system at equilibrium is a measure of the osmotic pressure and is calculated by using the equation,

π=cRT

where,

c is the molar concentration

Freezing point depression: The freezing point of the solution varies with the solute concentration.

Freezing point depression = ΔTfp= Kfp. msolute,where,Kfp=molal freezing point depression constant,msolute= molality of solute.

The number of moles of any substance can be determined using the equation

Numberofmole=GivenmassofthesubstanceMolarmass

### Explanation of Solution

Given,

R=0.082057L.atm K1mol1T=250C=(25+273.2)K=298.2K

Mass of the starch is 10.0g

Osmotic pressure,π=3.8mmHg

The osmotic pressure of the solution is,

π=cRT

Hence,

The concentration is calculated by,

Concentration,c=πRT=(3.8mmHg)(1atm760mmHg)(0.082057L.atmK1mol1)(298.2K)=2.0434×104mol/L

The amount of starch dissolved in one liter of water is,

2

(b)

Interpretation Introduction

Interpretation: The average molar mass of starch and freezing point of the solution is to be determined. To find out whether molar mass of starch can be obtained by using the depression of freezing point

Concept introduction:

Colligative properties: Properties of solutions which having influence on the concentration of the solute in it. Colligative properties are,

• Decrease in the vapor pressure
• Increase in the boiling point
• Decline in the freezing point
• Osmotic pressure

Osmotic pressure: The pressure created by the column of solution for the system at equilibrium is a measure of the osmotic pressure and is calculated by using the equation,

π=cRT

where,

c is the molar concentration

Freezing point depression: The freezing point of the solution varies with the solute concentration.

Freezing point depression = ΔTfp= Kfp. msolute,where,Kfp=molal freezing point depression constant,msolute= molality of solute.

The number of moles of any substance can be determined using the equation

Numberofmole=GivenmassofthesubstanceMolarmass

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