Chemistry for Today: General, Organic, and Biochemistry
Chemistry for Today: General, Organic, and Biochemistry
9th Edition
ISBN: 9781305960060
Author: Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen
Publisher: Cengage Learning
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Chapter 7, Problem 7.69E

Calculate the osmolarity for the following solutions:

a. A 0.25 M solution of KCl , a strong electrolyte

b. A solution containing 15.0 g of urea ( CH 4 N 2 O ) , a nonelectrolyte, per 500. mL

c. A solution containing 50.0 mL of ethylene glycol ( C 2 H 6 O 2 ) , a nonelectrolyte with a density of 1.11 g / mL , per 250. mL

Expert Solution
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Interpretation Introduction

(a)

Interpretation:

The osmolarity for a 0.25M solution of KCl, a strong electrolyte is to be calculated.

Concept introduction:

Solutes which give conducting solutions on dissolution are called electrolytes. Those which dissociate completely in the solution are known as strong electrolytes. Solutes which do not give conducting solutions are called nonelectrolytes. The properties which depend on the number of solute particles are known as colligative properties. Some of these properties are boiling point, freezing point and osmotic pressure.

Answer to Problem 7.69E

The osmolarity for a 0.25M solution of KCl, a strong electrolyte is 0.5M.

Explanation of Solution

The formula to calculate osmolarity is given below as,

Osmolarity=nM

Where,

n is the number of ions in the solution.

M is the molarity of solution.

Since KCl is a strong electrolyte, it will dissociate in the solution completely and the value of n for KCl is 2.

Substitute the value of n and M in the above equation as follows.

Osmolarity=nM=2×0.25M=0.5M

Conclusion

The osmolarity for a 0.25M solution of KCl, a strong electrolyte is 0.5M.

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation:

The osmolarity for a solution containing 15.0g of urea (CH4N2O), a nonelectrolyte, per 500.mL is to be calculated.

Concept introduction:

Solutes which give conducting solutions on dissolution are called electrolytes. Those which dissociate completely in the solution are known as strong electrolytes. Solutes which do not give conducting solutions are called nonelectrolytes. The properties which depend on the number of solute particles are known as colligative properties. Some of these properties are boiling point, freezing point and osmotic pressure.

Answer to Problem 7.69E

The osmolarity for a solution containing 15.0g of urea (CH4N2O), a nonelectrolyte, per 500.mL is 0.5M.

Explanation of Solution

The formula to calculate number of moles of solutes is given below as,

Molesofsolute=GivenmassofsoluteMolarmassofsolute

The molar mass of urea can be calculated as follows.

CH4N2O=(1×C)+(4×H)+(2×N)+(1×O)=(1×12)+(4×1)+(2×14)+(1×16)=60g/mol

Substitute the values in the above equation as follows.

MolesofCH4N2O=GivenmassofCH4N2OMolarmassofCH4N2O=15g60g/mol=0.25mol

The formula to calculate molarity is given below as,

M=molesofsolutelitersofsolution

Substitute the value of number of moles of solute and volume of solution in the above equation as follows.

M=molesofCH4N2OlitersofCH4N2Osolution=0.25mol500mL×1L1000mL=0.5M

The formula to calculate osmolarity is given below as,

Osmolarity=nM

Where,

n is the number of ions in the solution.

M is the molarity of solution.

Since urea is a nonelectrolyte, it will not dissociate in the solution and the value of n for urea is 1.

Substitute the value of n and M in the above equation as follows.

Osmolarity=nM=1×0.5M=0.5M

Conclusion

The osmolarity for a solution containing 15.0g of urea (CH4N2O), a nonelectrolyte, per 500.mL is 0.5M.

Expert Solution
Check Mark
Interpretation Introduction

(c)

Interpretation:

The osmolarity for a solution containing 50.0mL of ethylene glycol (C2H6O2), a nonelectrolyte with a density of 1.11g/mL, per 250.mL is to be calculated.

Concept introduction:

Solutes which give conducting solutions on dissolution are called electrolytes. Those which dissociate completely in the solution are known as strong electrolytes. Solutes which do not give conducting solutions are called nonelectrolytes. The properties which depend on the number of solute particles are known as colligative properties. Some of these properties are boiling point, freezing point and osmotic pressure.

Answer to Problem 7.69E

The osmolarity for a solution containing 50.0mL of ethylene glycol (C2H6O2), a nonelectrolyte with a density of 1.11g/mL, per 250.mL is 3.58M.

Explanation of Solution

The formula to calculate density is given below as,

d=mV

Where,

m is the mass of the substance.

V is the volume of the substance.

d is the density of the substance.

Thus, the mass of 50.0mL of ethylene glycol (C2H6O2) can be calculated as follows.

d=mVm=d×V=1.11g/mL×50.0mL=55.5g

The formula to calculate number of moles of solutes is given below as,

Molesofsolute=GivenmassofsoluteMolarmassofsolute

The molar mass of ethylene glycol can be calculated as follows.

C2H6O2=(2×C)+(6×H)+(2×O)=(2×12)+(6×1)+(2×16)=62g/mol

Substitute the values in the above equation as follows.

MolesofC2H6O2=GivenmassofC2H6O2MolarmassofC2H6O2=55.5g62g/mol=0.895mol

The formula to calculate molarity is given below as,

M=molesofsolutelitersofsolution

Substitute the value of number of moles of solute and volume of solution in the above equation as follows.

M=molesofC2H6O2litersofC2H6O2solution=0.895mol250mL×1L1000mL=3.58M

The formula to calculate osmolarity is given below as,

Osmolarity=nM

Where,

n is the number of ions in the solution.

M is the molarity of solution.

Since ethylene glycol is a nonelectrolyte, it will not dissociate in the solution and the value of n for ethylene glycol is 1.

Substitute the value of n and M in the above equation as follows.

Osmolarity=nM=1×3.58M=3.58M

Conclusion

The osmolarity for a solution containing 50.0mL of ethylene glycol (C2H6O2), a nonelectrolyte with a density of 1.11g/mL, per 250.mL is 3.58M.

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Chapter 7 Solutions

Chemistry for Today: General, Organic, and Biochemistry

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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