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 7.43E

Explain how you would prepare the following solutions using pure solute and water. Assume water has a density of 1 .00 g / mL .

a. 250 mL of a 2 .00 M NaOH solution.

b. 500 mL of a 40.0 % ( v / v ) alcohol solution ( C 2 H 5 OH )

c. 100 mL of 15.0 % ( w / v ) glycerol solution. Glycerol is a liquid with a density of 1 .26 g / mL . Describe two ways to measure out the amount of glycerol needed.

d. Approximately 50 mL of a normal saline solution, 0 .89% ( w / w ) NaCl .

Expert Solution
Check Mark
Interpretation Introduction

(a)

Interpretation:

The preparation of the given solutions using pure solute and water is to be explained.

Concept Introduction:

The number of moles is calculated by the formula,

Molarity=NumberofmolesVolume(mL)×1000mL1L

Answer to Problem 7.43E

2.00g of NaOH is mixed with 250mL of water to obtain the final solution.

Explanation of Solution

The number of moles of NaOH is calculated by the formula,

Molarity=NumberofmolesVolume(mL)×1000mL1L

The above formula can be written as follows:

Numberofmoles=Molarity×Volume(mL)×1L1000mL … (1)

The given volume and molarity is 250mL and 2.00M respectively.

Substitute the volume and molarity in the equation (1).

Numberofmoles=0.200M×250mL×1L1000mLNumberofmoles=0.05mol

Thus, the number of moles of NaOH is 0.05mol.

The amount of NaOH is calculated by the formula,

Moles=GivenmassMolarmass

The molar mass of NaOH is 40g/mol.

Substitute the value of molar mass in the given formula.

0.05mol=AmountofNaOH40g/molAmountofNaOH=0.05mol×40g/mol=2.00g

Thus, the amount of NaOH is 2.00g.

Hence, 2.00g of NaOH is mixed with 250mL of water to obtain the final solution.

Conclusion

2.00g of NaOH is mixed with 250mL of water to obtain the final solution.

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation:

The preparation of the given solutions using pure solute and water is to be explained.

Concept Introduction:

The concentration of the solution in %(v/v) is given by the formula,

%(v/v)=solutevolumesolutionvolume×100

Answer to Problem 7.43E

200mL of alcohol solution is mixed with 500mL of water to obtain the final solution.

Explanation of Solution

The concentration of the solution in %(v/v) is given by the formula,

%(v/v)=solutevolumesolutionvolume×100 … (1)

The given value of %(v/v) is 40.0%(v/v). The volume of solution is 500mL.

Substitute the value of %(v/v) and volume of solution in equation (1).

40.0=solutevolume 500mL×100solutevolume =500 ×40.0100=200mL

Hence, the volume of solute is 200mL.

Therefore, 200mL of alcohol solution is mixed with 500mL of water to obtain the final solution.

Conclusion

200mL of alcohol solution is mixed with 500mL of water to obtain the final solution.

Expert Solution
Check Mark
Interpretation Introduction

(c)

Interpretation:

The preparation of the given solutions using pure solute and water is to be explained. The two ways to measure out the amount of glycerol is to be described.

Concept Introduction:

The concentration of the solution in %(w/v) is given by the formula,

%(w/v)=gramsofsolutemillilitersofsolution×100

Answer to Problem 7.43E

15.0g of glycerol is mixed with 100mL of water to obtain the final solution. 11.90mL of glycerol along with remaining amount of water, is added to obtain the final solution.

Explanation of Solution

The concentration of the solution in %(w/v) is given by the formula,

%(w/v)=gramsofsolutemillilitersofsolution×100 …(1)

The given value of %(w/v) is 15.0%(w/v). The volume of glycerol solution is 100mL.

Substitute the value of %(w/v) and volume in equation (1).

15.0=gramsofsolute100mL×100gramsofsolute=15.0×100mL100=15.0g

Hence, 15.0g of glycerol is mixed with 100mL of water to obtain the final solution.

The density of glycerol is 1.26g/mL.

The density is given by the formula,

Density=MassofsolutionVolumeofsolution

Substitute the density and mass of glycerol and in the above formula.

1.26g/mL=15.0gvolumeofglycerolvolumeofglycerol=15.0g1.26g/mL=11.90mL

Hence, 11.90mL of glycerol along with remaining amount of water, is added to obtain the final solution.

Conclusion

15.0g of glycerol is mixed with 100mL of water to obtain the final solution. 11.90mL of glycerol along with remaining amount of water, is added to obtain the final solution.

Expert Solution
Check Mark
Interpretation Introduction

(d)

Interpretation:

The preparation of the given solutions using pure solute and water is to be explained.

Concept Introduction:

The concentration of the solution in %(w/w) is given by the formula,

%(w/w)=solutemasssolutionmass×100

Answer to Problem 7.43E

0.445g of sodium chloride is mixed with 50mL of water to obtain the final solution.

Explanation of Solution

The concentration of the solution in %(w/w) is given by the formula,

%(w/w)=solutemasssolutionmass×100 … (1)

The given value of %(w/w) is 0.89%(w/w). The density of water is 1g/mL. Thus, the mass of water in 50mL will be 50g. Therefore, the mass of the solution is 50g.

Substitute the value of %(w/w) and mass of solution in equation (1).

0.89=solutemass 50g×100solutemass =50g ×0.89100=0.445g

Hence, the mass of solute is 0.445g.

Therefore, 0.445g of sodium chloride is mixed with 50mL of water to obtain the final solution.

Conclusion

0.445g of sodium chloride is mixed with 50mL of water to obtain the final solution.

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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