   # An aqueous solution is 1.00% NaCl by mass and has a density of 1.071 g/cm 3 at 25°C. The observed osmotic pressure of this solution is 7.83 atm at 25°C. a. What fraction of the moles of NaCl in this solution exist as ion pairs? b. Calculate the freezing point that would be observed for this solution. ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 10, Problem 122CP
Textbook Problem
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## An aqueous solution is 1.00% NaCl by mass and has a density of 1.071 g/cm3 at 25°C. The observed osmotic pressure of this solution is 7.83 atm at 25°C.a. What fraction of the moles of NaCl in this solution exist as ion pairs?b. Calculate the freezing point that would be observed for this solution.

Interpretation Introduction

Interpretation:

The fraction of moles of NaCl existing as ion pairs and the freezing point of the solution has to be calculated.

Concept Introduction:

Mole fraction of a compound can be defined as the number of moles of a substance to the total number of moles present in them. The mole fraction can be calculated by,

Molefractionofcompound=Numberofmoles(inmol)Totalnumberofmoles(inmol)

Freezing point of solution can be calculated from the equation,

ΔTf=Kfm

ΔTf = change in freezing point

Kf = molal depression freezing point constant

m   = moles of solute particles

### Explanation of Solution

Record the given info

Mass percentage of NaCl = 1.00%

Density of NaCl at 25°C   = 1.071gcm-3

Osmotic pressure at 25°C = 7.83atm

To calculate the molarity of the solution

From the osmotic pressure equation,

π=iMRTiM=πRT=7.83atm0.08206Latm/K.mol×298K=0.320molL-1

Molarity of solution = 0.320M

To calculate the moles of NaCl added to the solution

Assume 1.00L of the solution,

Total moles of solute particles         = molNa++molCl-+molNaCl=0.320mol

Mass of the solution                        = 1000mL×1.071gmL=1071gsolution

Mass of NaCl in solution                = 0.0100×1071g=10.7g

Moles of NaCl added to the solution = 10.71mol58.44g=0.183mol

To calculate the mole fraction of NaCl existing as ion pairs

Let x be moles of undissociated moles of NaCl = mol ion pairs

Moles of solute particles = 0

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