Chemistry for Today: General  Organic  and Biochemistry
Chemistry for Today: General Organic and Biochemistry
9th Edition
ISBN: 9781337514576
Author: Seager
Publisher: Cengage
Question
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Chapter 7, Problem 7.42E
Interpretation Introduction

(a)

Interpretation:

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

Concept Introduction:

The number of moles is calculated by the formula,

Molarity=NumberofmolesVolume(mL)×1000mL1L

Expert Solution
Check Mark

Answer to Problem 7.42E

4.26g of Na2SO4 is mixed with 200mL of water to obtain the final solution.

Explanation of Solution

The number of moles of Na2SO4 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 200mL and 0.150M respectively.

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

Numberofmoles=0.150M×200mL×1L1000mLNumberofmoles=0.03mol

Thus, the number of moles of Na2SO4 is 0.03mol.

The amount of Na2SO4 is calculated by the formula,

Moles=GivenmassMolarmass

The molar mass of Na2SO4 is 142.04g/mol.

Substitute the value of molar mass in the given formula.

0.03mol=AmountofNa2SO4142.04g/molAmountofNa2SO4=0.03mol×142.04g/mol=4.26g

Thus, the amount of Na2SO4 is 4.26g.

Hence, 4.26g of Na2SO4 is mixed with 200mL of water to obtain the final solution.

Conclusion

4.26g of Na2SO4 is mixed with 200mL of water to obtain the final solution.

Interpretation Introduction

(b)

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

Expert Solution
Check Mark

Answer to Problem 7.42E

11.8g of Zn(NO3)2 is mixed with 250mL of water to obtain the final solution.

Explanation of Solution

The number of moles of Zn(NO3)2 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 0.250M respectively.

Substitute the volume and molarity in equation (1).

Numberofmoles=0.250M×250mL×1L1000mLNumberofmoles=0.0625mol

Thus, the number of moles of Zn(NO3)2 is 0.0625mol.

The amount of Zn(NO3)2 is calculated by the formula,

Moles=GivenmassMolarmass

The molar mass of Zn(NO3)2 is 189.4g/mol.

Substitute the value of molar mass in the given formula.

0.0625mol=AmountofZn(NO3)2189.4g/molAmountofZn(NO3)2=0.0625mol×189.4g/mol=11.8g

Thus, the amount of Zn(NO3)2 is 11.8g.

Hence, 11.8g of Zn(NO3)2 is mixed with 250mL of water to obtain the final solution.

Conclusion

11.8g of Zn(NO3)2 is mixed with 250mL of water to obtain the final solution.

Interpretation Introduction

(c)

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

Expert Solution
Check Mark

Answer to Problem 7.42E

3.4g of sodium chloride is mixed with 146.6mL of water to obtain the final solution.

Explanation of Solution

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

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

The given value of %(w/w) is 2.25%(w/w). The mass of solution is 150g.

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

2.25=solutemass 150g×100solutemass=150×2.25100=3.4g

Hence, the mass of solute is 3.4g.

Mass of solvent is calculated as follows:

Massofsolution=Massofsolute+Massofsolvent

Substitute the mass of solute and solution in above formula.

150g=3.4g+MassofsolventMassofsolvent=(1503.4)g=146.6g

Thus, the mass of solvent (water) is 146.6g. The density of water is 1.00g/mL. Hence, the volume of water is 146.6mL.

Therefore, 3.4g of sodium chloride is mixed with 146.6mL of water to obtain the final solution.

Conclusion

3.4g of sodium chloride is mixed with 146.6mL of water to obtain the final solution.

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/v) is given by the formula,

%(w/v)=gramsofsolutemillilitersofsolution×100

Expert Solution
Check Mark

Answer to Problem 7.42E

0.94g of KCl is mixed with 125mL of water 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 0.75%(w/v). The volume of KCl solution is 125mL.

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

0.75=gramsofsolute125mL×100gramsofsolute=0.75×125mL100=0.94g

Hence, 0.94g of KCl is mixed with 125mL of water to obtain the final solution.

Conclusion

0.94g of KCl is mixed with 125mL 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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