# A salt solution has an osmotic pressure of 16 atmospheres at 22°C. What is the freezing point of this solution? What assumptions must be made to solve this problem?

### General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
Publisher: Cengage Learning
ISBN: 9781305580343

### General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
Publisher: Cengage Learning
ISBN: 9781305580343

#### Solutions

Chapter
Section
Chapter 12, Problem 12.110QP
Textbook Problem

## A salt solution has an osmotic pressure of 16 atmospheres at 22°C. What is the freezing point of this solution? What assumptions must be made to solve this problem?

Expert Solution
Interpretation Introduction

Interpretation:

Freezing point of a solution at given Osmotic pressure and temperature has to be calculated.

Concept Introduction:

In the process of osmosis, the solvent molecules pass through a semi - permeable membrane from less concentrated solution to more concentrated solution.  The pressure that has to be applied to prevent the flow of solvent molecules is called osmotic pressure.  It is expressed as,

π = MRT

Where,

π = osmotic pressureM = Molar concentrationR =  Universal gas constantT =   Temperature

Freezing point of a liquid is defined as the temperature at which liquid becomes solid and the two phases remain in equilibrium with each other.

Freezing point of a substance can be determined by the formula,

ΔTf = i×Kfm

Where,

ΔTf = depression in freezing pointKf  = cryoscopic constantm    = molality of the solution; i = van't Hoff factor

### Explanation of Solution

Given that osmotic pressure of solution is 16 atm at 22°C .

As we know,

Osmotic pressure of glucose solution at 22°C(= 295.2 K) is,

π= MRT = 16 atm

Rearrange the above expression to determine Molarity,

π  = MRTM πRT = 16 atm(0.0821 L.atm/K.mol)× (295.2 K) = 0.660 mol/L = 0.660 M

The molar concentration of the solution indicates it is very dilute solution.  In case of very dilute solutions the molality and molarity don’t vary much that the values are almost equal.  Hence we can assume that molality and molarity are approximately similar and the solute exhibits ideal behavior.  The freezing point is calculated by determining depression in freezing point.  Here water is the solvent.  We know Kf for water is 1.86°C/m

ΔTf = Kf×m = 1

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