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.

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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, Mole fraction of compound= Number of moles(in mol) Total number of moles(in mol) Freezing point of solution can be calculated from the equation, ΔT f =K f m ΔT f = change in freezing point K f = molal depression freezing point constant m = moles of solute particles Explanation Record the given info Mass percentage of NaCl = 1 .00% Density of NaCl at 25°C = 1 .071 g cm -3 Osmotic pressure at 25°C = 7 .83 atm To calculate the molarity of the solution From the osmotic pressure equation, π=iMRT iM= π RT = 7 .83 atm 0 .08206 L atm/K .mol×298K =0 .320 mol L -1 Molarity of solution = 0 .320 M To calculate the moles of NaCl added to the solution Assume 1 .00 L of the solution, Total moles of solute particles = mol Na + +mol Cl - +mol NaCl=0 .320 mol Mass of the solution = 1000 mL× 1 .071 g mL =1071 g solution Mass of NaCl in solution = 0 .0100×1071 g =10 .7 g Moles of NaCl added to the solution = 10 .7 g× 1 mol 58 .44 g =0 .183 mol 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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Chemistry: An Atoms First Approach

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Chemistry: An Atoms First Approach

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Steven S. Zumdahl + 1 other

Publisher: Cengage Learning

ISBN: 9781305079243

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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,

Mole fraction of compound=Number of moles(in mol)Total number of moles(in mol)

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 = 1.071 g cm-3

Osmotic pressure at 25°C = 7.83 atm

To calculate the molarity of the solution

From the osmotic pressure equation,

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

Molarity of solution = 0.320 M

To calculate the moles of NaCl added to the solution

Assume 1.00 L of the solution,

Total moles of solute particles = mol Na++mol Cl-+mol NaCl=0.320 mol

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

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

Moles of NaCl added to the solution = 10.7 g×1 mol58.44 g=0.183 mol

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

Chemistry: An Atoms First Approach

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