The Normality and equivalent mass of given following solution has to be calculated. Concept Introduction: Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula, Normality(N)= Number of gram equivalent(in g) Volume of solution(in L)

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 11, Problem 38E

(a)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(a)

Expert Solution

Answer to Problem 38E

The normality of $HCl$ is $0.25N$ $Equivalent mass = molar mass of HCl =36.46 g$

Explanation of Solution

Record the given data

Gram equivalent of $HCl$ = $0.25$

Molarity of $HCl$ = $0.25 M$

To calculate the normality of $HCl$

$HCl$ has one acidic protons because it is a monoprotic acid Hence, its molarity can be one time

Therefore, the normality of $HCl$ is given by,

$Normality = 0.25 mol HClL×1 equivalentmol HCl=0.25 equivalentsLEquivalent mass = molar mass of HCl =36.46 g$

(b)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(b)

Expert Solution

Answer to Problem 38E

The normality of $H2SO4$ is $0.21N$

Explanation of Solution

Record the given data

Gram equivalent of $H2SO4$ = $0.21$

Molarity of $H2SO4$ = $0.105 M$

The gram equivalent and molarity of $H2SO4$ are recorded as shown above.

To calculate the normality of $H2SO4$

$H2SO4$ has two acidic protons because it is a diprotic acid Hence, its molarity can be two times

Therefore, the normality of $H2SO4$ is given by,

$Normality = 0.105 mol H2SO4L×2 equivalentmol H2SO4=0.210 equivalentsLEquivalent mass = 12(molar mass of H2SO4) =12(98.09)=49.05 g$

(c)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(c)

Expert Solution

Answer to Problem 38E

The normality of $H3PO4$ is $0.16N$

Explanation of Solution

Record the given data

Gram equivalent of $H3PO4$ = $0.16$

Molarity of $H3PO4$ = $5.3×10-2 M$

To calculate the normality of $H3PO4$

$H3PO4$ has three acidic protons because it is a diprotic acid Hence, its molarity can be three times

Therefore, the normality of $H3PO4$ is given by,

$Normality = 5.3× 10-2 mol H3PO4L×3 equivalentmol H3PO4=0.16 equivalentsLEquivalent mass = 13(molar mass of H3PO4) =13(97.09)=32.66 g$

(d)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(d)

Expert Solution

Answer to Problem 38E

The normality of $NaOH$ is $0.134N$

Explanation of Solution

Record the given data

Gram equivalent of $NaOH$ = $0.134$

Molarity of $NaOH$ = $0.134 M$

To calculate the normality of $NaOH$

$NaOH$ has mono acidic protons because it is a monoprotic acid Hence, its molarity can be one times

Therefore, the normality of $NaOH$ is given by,

$Normality = 0.134 mol NaOHL×1 equivalentmol NaOH=0.134 equivalentsLEquivalent mass = molar mass of NaOH 40 g$

(e)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(e)

Expert Solution

Answer to Problem 38E

The normality of $Ca(OH)2$ is $0.002605 N$

Explanation of Solution

Record the given data

Gram equivalent of $Ca(OH)2$ = $0.0104$

Molarity of $Ca(OH)2$ = $0.00521 M$

To calculate the normality of $Ca(OH)2$

$NaOH$ has di acidic protons because it is a diprotic acid Hence, its molarity can be two times

Therefore, the normality of $Ca(OH)2$ is given by,

$Normality = 0.00521 mol Ca(OH)2L×2 equivalentmol Ca(OH)2=0.0104 equivalentsLEquivalent mass = 12(molar mass of Ca(OH)2)=12(74.10)=37.05 g$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Please help me calculate the undiluted samples ppm concentration. My calculations were 280.11 ppm. Please see if I did my math correctly using the following standard curve. Link: https://mnscu-my.sharepoint.com/:x:/g/personal/vi2163ss_go_minnstate_edu/EVSJL_W0qrxMkUjK2J3xMUEBHDu0UM1vPKQ-bc9HTcYXDQ?e=hVuPC4

Provide an IUPAC name for each of the compounds shown. (Specify (E)/(Z) stereochemistry, if relevant, for straight chain alkenes only. Pay attention to commas, dashes, etc.) H₁₂C C(CH3)3 C=C H3C CH3 CH3CH2CH CI CH3 Submit Answer Retry Entire Group 2 more group attempts remaining Previous Next

Arrange the following compounds / ions in increasing nucleophilicity (least to most nucleophilic) CH3NH2 CH3C=C: CH3COO 1 2 3 5 Multiple Choice 1 point 1, 2, 3 2, 1, 3 3, 1, 2 2, 3, 1 The other answers are not correct 0000

Answer 2

Question

Chapter 11, Problem 38E

(a)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(a)

Expert Solution

Answer to Problem 38E

The normality of $HCl$ is $0.25N$ $Equivalent mass = molar mass of HCl =36.46 g$

Explanation of Solution

Record the given data

Gram equivalent of $HCl$ = $0.25$

Molarity of $HCl$ = $0.25 M$

To calculate the normality of $HCl$

$HCl$ has one acidic protons because it is a monoprotic acid Hence, its molarity can be one time

Therefore, the normality of $HCl$ is given by,

$Normality = 0.25 mol HClL×1 equivalentmol HCl=0.25 equivalentsLEquivalent mass = molar mass of HCl =36.46 g$

(b)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(b)

Expert Solution

Answer to Problem 38E

The normality of $H2SO4$ is $0.21N$

Explanation of Solution

Record the given data

Gram equivalent of $H2SO4$ = $0.21$

Molarity of $H2SO4$ = $0.105 M$

The gram equivalent and molarity of $H2SO4$ are recorded as shown above.

To calculate the normality of $H2SO4$

$H2SO4$ has two acidic protons because it is a diprotic acid Hence, its molarity can be two times

Therefore, the normality of $H2SO4$ is given by,

$Normality = 0.105 mol H2SO4L×2 equivalentmol H2SO4=0.210 equivalentsLEquivalent mass = 12(molar mass of H2SO4) =12(98.09)=49.05 g$

(c)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(c)

Expert Solution

Answer to Problem 38E

The normality of $H3PO4$ is $0.16N$

Explanation of Solution

Record the given data

Gram equivalent of $H3PO4$ = $0.16$

Molarity of $H3PO4$ = $5.3×10-2 M$

To calculate the normality of $H3PO4$

$H3PO4$ has three acidic protons because it is a diprotic acid Hence, its molarity can be three times

Therefore, the normality of $H3PO4$ is given by,

$Normality = 5.3× 10-2 mol H3PO4L×3 equivalentmol H3PO4=0.16 equivalentsLEquivalent mass = 13(molar mass of H3PO4) =13(97.09)=32.66 g$

(d)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(d)

Expert Solution

Answer to Problem 38E

The normality of $NaOH$ is $0.134N$

Explanation of Solution

Record the given data

Gram equivalent of $NaOH$ = $0.134$

Molarity of $NaOH$ = $0.134 M$

To calculate the normality of $NaOH$

$NaOH$ has mono acidic protons because it is a monoprotic acid Hence, its molarity can be one times

Therefore, the normality of $NaOH$ is given by,

$Normality = 0.134 mol NaOHL×1 equivalentmol NaOH=0.134 equivalentsLEquivalent mass = molar mass of NaOH 40 g$

(e)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(e)

Expert Solution

Answer to Problem 38E

The normality of $Ca(OH)2$ is $0.002605 N$

Explanation of Solution

Record the given data

Gram equivalent of $Ca(OH)2$ = $0.0104$

Molarity of $Ca(OH)2$ = $0.00521 M$

To calculate the normality of $Ca(OH)2$

$NaOH$ has di acidic protons because it is a diprotic acid Hence, its molarity can be two times

Therefore, the normality of $Ca(OH)2$ is given by,

$Normality = 0.00521 mol Ca(OH)2L×2 equivalentmol Ca(OH)2=0.0104 equivalentsLEquivalent mass = 12(molar mass of Ca(OH)2)=12(74.10)=37.05 g$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 11, Problem 38E

(a)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(a)

Expert Solution

Answer to Problem 38E

The normality of $HCl$ is $0.25N$ $Equivalent mass = molar mass of HCl =36.46 g$

Explanation of Solution

Record the given data

Gram equivalent of $HCl$ = $0.25$

Molarity of $HCl$ = $0.25 M$

To calculate the normality of $HCl$

$HCl$ has one acidic protons because it is a monoprotic acid Hence, its molarity can be one time

Therefore, the normality of $HCl$ is given by,

$Normality = 0.25 mol HClL×1 equivalentmol HCl=0.25 equivalentsLEquivalent mass = molar mass of HCl =36.46 g$

(b)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(b)

Expert Solution

Answer to Problem 38E

The normality of $H2SO4$ is $0.21N$

Explanation of Solution

Record the given data

Gram equivalent of $H2SO4$ = $0.21$

Molarity of $H2SO4$ = $0.105 M$

The gram equivalent and molarity of $H2SO4$ are recorded as shown above.

To calculate the normality of $H2SO4$

$H2SO4$ has two acidic protons because it is a diprotic acid Hence, its molarity can be two times

Therefore, the normality of $H2SO4$ is given by,

$Normality = 0.105 mol H2SO4L×2 equivalentmol H2SO4=0.210 equivalentsLEquivalent mass = 12(molar mass of H2SO4) =12(98.09)=49.05 g$

(c)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(c)

Expert Solution

Answer to Problem 38E

The normality of $H3PO4$ is $0.16N$

Explanation of Solution

Record the given data

Gram equivalent of $H3PO4$ = $0.16$

Molarity of $H3PO4$ = $5.3×10-2 M$

To calculate the normality of $H3PO4$

$H3PO4$ has three acidic protons because it is a diprotic acid Hence, its molarity can be three times

Therefore, the normality of $H3PO4$ is given by,

$Normality = 5.3× 10-2 mol H3PO4L×3 equivalentmol H3PO4=0.16 equivalentsLEquivalent mass = 13(molar mass of H3PO4) =13(97.09)=32.66 g$

(d)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(d)

Expert Solution

Answer to Problem 38E

The normality of $NaOH$ is $0.134N$

Explanation of Solution

Record the given data

Gram equivalent of $NaOH$ = $0.134$

Molarity of $NaOH$ = $0.134 M$

To calculate the normality of $NaOH$

$NaOH$ has mono acidic protons because it is a monoprotic acid Hence, its molarity can be one times

Therefore, the normality of $NaOH$ is given by,

$Normality = 0.134 mol NaOHL×1 equivalentmol NaOH=0.134 equivalentsLEquivalent mass = molar mass of NaOH 40 g$

(e)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(e)

Expert Solution

Answer to Problem 38E

The normality of $Ca(OH)2$ is $0.002605 N$

Explanation of Solution

Record the given data

Gram equivalent of $Ca(OH)2$ = $0.0104$

Molarity of $Ca(OH)2$ = $0.00521 M$

To calculate the normality of $Ca(OH)2$

$NaOH$ has di acidic protons because it is a diprotic acid Hence, its molarity can be two times

Therefore, the normality of $Ca(OH)2$ is given by,

$Normality = 0.00521 mol Ca(OH)2L×2 equivalentmol Ca(OH)2=0.0104 equivalentsLEquivalent mass = 12(molar mass of Ca(OH)2)=12(74.10)=37.05 g$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 11, Problem 38E

(a)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(a)

Expert Solution

Answer to Problem 38E

The normality of $HCl$ is $0.25N$ $Equivalent mass = molar mass of HCl =36.46 g$

Explanation of Solution

Record the given data

Gram equivalent of $HCl$ = $0.25$

Molarity of $HCl$ = $0.25 M$

To calculate the normality of $HCl$

$HCl$ has one acidic protons because it is a monoprotic acid Hence, its molarity can be one time

Therefore, the normality of $HCl$ is given by,

$Normality = 0.25 mol HClL×1 equivalentmol HCl=0.25 equivalentsLEquivalent mass = molar mass of HCl =36.46 g$

(b)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(b)

Expert Solution

Answer to Problem 38E

The normality of $H2SO4$ is $0.21N$

Explanation of Solution

Record the given data

Gram equivalent of $H2SO4$ = $0.21$

Molarity of $H2SO4$ = $0.105 M$

The gram equivalent and molarity of $H2SO4$ are recorded as shown above.

To calculate the normality of $H2SO4$

$H2SO4$ has two acidic protons because it is a diprotic acid Hence, its molarity can be two times

Therefore, the normality of $H2SO4$ is given by,

$Normality = 0.105 mol H2SO4L×2 equivalentmol H2SO4=0.210 equivalentsLEquivalent mass = 12(molar mass of H2SO4) =12(98.09)=49.05 g$

(c)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(c)

Expert Solution

Answer to Problem 38E

The normality of $H3PO4$ is $0.16N$

Explanation of Solution

Record the given data

Gram equivalent of $H3PO4$ = $0.16$

Molarity of $H3PO4$ = $5.3×10-2 M$

To calculate the normality of $H3PO4$

$H3PO4$ has three acidic protons because it is a diprotic acid Hence, its molarity can be three times

Therefore, the normality of $H3PO4$ is given by,

$Normality = 5.3× 10-2 mol H3PO4L×3 equivalentmol H3PO4=0.16 equivalentsLEquivalent mass = 13(molar mass of H3PO4) =13(97.09)=32.66 g$

(d)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(d)

Expert Solution

Answer to Problem 38E

The normality of $NaOH$ is $0.134N$

Explanation of Solution

Record the given data

Gram equivalent of $NaOH$ = $0.134$

Molarity of $NaOH$ = $0.134 M$

To calculate the normality of $NaOH$

$NaOH$ has mono acidic protons because it is a monoprotic acid Hence, its molarity can be one times

Therefore, the normality of $NaOH$ is given by,

$Normality = 0.134 mol NaOHL×1 equivalentmol NaOH=0.134 equivalentsLEquivalent mass = molar mass of NaOH 40 g$

(e)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(e)

Expert Solution

Answer to Problem 38E

The normality of $Ca(OH)2$ is $0.002605 N$

Explanation of Solution

Record the given data

Gram equivalent of $Ca(OH)2$ = $0.0104$

Molarity of $Ca(OH)2$ = $0.00521 M$

To calculate the normality of $Ca(OH)2$

$NaOH$ has di acidic protons because it is a diprotic acid Hence, its molarity can be two times

Therefore, the normality of $Ca(OH)2$ is given by,

$Normality = 0.00521 mol Ca(OH)2L×2 equivalentmol Ca(OH)2=0.0104 equivalentsLEquivalent mass = 12(molar mass of Ca(OH)2)=12(74.10)=37.05 g$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 11, Problem 38E

(a)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(a)

Expert Solution

Answer to Problem 38E

The normality of $HCl$ is $0.25N$ $Equivalent mass = molar mass of HCl =36.46 g$

Explanation of Solution

Record the given data

Gram equivalent of $HCl$ = $0.25$

Molarity of $HCl$ = $0.25 M$

To calculate the normality of $HCl$

$HCl$ has one acidic protons because it is a monoprotic acid Hence, its molarity can be one time

Therefore, the normality of $HCl$ is given by,

$Normality = 0.25 mol HClL×1 equivalentmol HCl=0.25 equivalentsLEquivalent mass = molar mass of HCl =36.46 g$

(b)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(b)

Expert Solution

Answer to Problem 38E

The normality of $H2SO4$ is $0.21N$

Explanation of Solution

Record the given data

Gram equivalent of $H2SO4$ = $0.21$

Molarity of $H2SO4$ = $0.105 M$

The gram equivalent and molarity of $H2SO4$ are recorded as shown above.

To calculate the normality of $H2SO4$

$H2SO4$ has two acidic protons because it is a diprotic acid Hence, its molarity can be two times

Therefore, the normality of $H2SO4$ is given by,

$Normality = 0.105 mol H2SO4L×2 equivalentmol H2SO4=0.210 equivalentsLEquivalent mass = 12(molar mass of H2SO4) =12(98.09)=49.05 g$

(c)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(c)

Expert Solution

Answer to Problem 38E

The normality of $H3PO4$ is $0.16N$

Explanation of Solution

Record the given data

Gram equivalent of $H3PO4$ = $0.16$

Molarity of $H3PO4$ = $5.3×10-2 M$

To calculate the normality of $H3PO4$

$H3PO4$ has three acidic protons because it is a diprotic acid Hence, its molarity can be three times

Therefore, the normality of $H3PO4$ is given by,

$Normality = 5.3× 10-2 mol H3PO4L×3 equivalentmol H3PO4=0.16 equivalentsLEquivalent mass = 13(molar mass of H3PO4) =13(97.09)=32.66 g$

(d)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(d)

Expert Solution

Answer to Problem 38E

The normality of $NaOH$ is $0.134N$

Explanation of Solution

Record the given data

Gram equivalent of $NaOH$ = $0.134$

Molarity of $NaOH$ = $0.134 M$

To calculate the normality of $NaOH$

$NaOH$ has mono acidic protons because it is a monoprotic acid Hence, its molarity can be one times

Therefore, the normality of $NaOH$ is given by,

$Normality = 0.134 mol NaOHL×1 equivalentmol NaOH=0.134 equivalentsLEquivalent mass = molar mass of NaOH 40 g$

(e)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(e)

Expert Solution

Answer to Problem 38E

The normality of $Ca(OH)2$ is $0.002605 N$

Explanation of Solution

Record the given data

Gram equivalent of $Ca(OH)2$ = $0.0104$

Molarity of $Ca(OH)2$ = $0.00521 M$

To calculate the normality of $Ca(OH)2$

$NaOH$ has di acidic protons because it is a diprotic acid Hence, its molarity can be two times

Therefore, the normality of $Ca(OH)2$ is given by,

$Normality = 0.00521 mol Ca(OH)2L×2 equivalentmol Ca(OH)2=0.0104 equivalentsLEquivalent mass = 12(molar mass of Ca(OH)2)=12(74.10)=37.05 g$

Answer 6

Chapter 11, Problem 38E

(a)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(a)

Expert Solution

Answer to Problem 38E

The normality of $HCl$ is $0.25N$ $Equivalent mass = molar mass of HCl =36.46 g$

Explanation of Solution

Record the given data

Gram equivalent of $HCl$ = $0.25$

Molarity of $HCl$ = $0.25 M$

To calculate the normality of $HCl$

$HCl$ has one acidic protons because it is a monoprotic acid Hence, its molarity can be one time

Therefore, the normality of $HCl$ is given by,

$Normality = 0.25 mol HClL×1 equivalentmol HCl=0.25 equivalentsLEquivalent mass = molar mass of HCl =36.46 g$

Answer 7

Chapter 11, Problem 38E

(a)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

Answer 8

(a)

Expert Solution

Answer to Problem 38E

The normality of $HCl$ is $0.25N$ $Equivalent mass = molar mass of HCl =36.46 g$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Record the given data

Gram equivalent of $HCl$ = $0.25$

Molarity of $HCl$ = $0.25 M$

To calculate the normality of $HCl$

$HCl$ has one acidic protons because it is a monoprotic acid Hence, its molarity can be one time

Therefore, the normality of $HCl$ is given by,

$Normality = 0.25 mol HClL×1 equivalentmol HCl=0.25 equivalentsLEquivalent mass = molar mass of HCl =36.46 g$

Answer 13

(b)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(b)

Expert Solution

Answer to Problem 38E

The normality of $H2SO4$ is $0.21N$

Explanation of Solution

Record the given data

Gram equivalent of $H2SO4$ = $0.21$

Molarity of $H2SO4$ = $0.105 M$

The gram equivalent and molarity of $H2SO4$ are recorded as shown above.

To calculate the normality of $H2SO4$

$H2SO4$ has two acidic protons because it is a diprotic acid Hence, its molarity can be two times

Therefore, the normality of $H2SO4$ is given by,

$Normality = 0.105 mol H2SO4L×2 equivalentmol H2SO4=0.210 equivalentsLEquivalent mass = 12(molar mass of H2SO4) =12(98.09)=49.05 g$

Answer 14

(b)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

Answer 15

(b)

Expert Solution

Answer to Problem 38E

The normality of $H2SO4$ is $0.21N$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Record the given data

Gram equivalent of $H2SO4$ = $0.21$

Molarity of $H2SO4$ = $0.105 M$

The gram equivalent and molarity of $H2SO4$ are recorded as shown above.

To calculate the normality of $H2SO4$

$H2SO4$ has two acidic protons because it is a diprotic acid Hence, its molarity can be two times

Therefore, the normality of $H2SO4$ is given by,

$Normality = 0.105 mol H2SO4L×2 equivalentmol H2SO4=0.210 equivalentsLEquivalent mass = 12(molar mass of H2SO4) =12(98.09)=49.05 g$

Answer 20

(c)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(c)

Expert Solution

Answer to Problem 38E

The normality of $H3PO4$ is $0.16N$

Explanation of Solution

Record the given data

Gram equivalent of $H3PO4$ = $0.16$

Molarity of $H3PO4$ = $5.3×10-2 M$

To calculate the normality of $H3PO4$

$H3PO4$ has three acidic protons because it is a diprotic acid Hence, its molarity can be three times

Therefore, the normality of $H3PO4$ is given by,

$Normality = 5.3× 10-2 mol H3PO4L×3 equivalentmol H3PO4=0.16 equivalentsLEquivalent mass = 13(molar mass of H3PO4) =13(97.09)=32.66 g$

Answer 21

(c)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

Answer 22

(c)

Expert Solution

Answer to Problem 38E

The normality of $H3PO4$ is $0.16N$

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Record the given data

Gram equivalent of $H3PO4$ = $0.16$

Molarity of $H3PO4$ = $5.3×10-2 M$

To calculate the normality of $H3PO4$

$H3PO4$ has three acidic protons because it is a diprotic acid Hence, its molarity can be three times

Therefore, the normality of $H3PO4$ is given by,

$Normality = 5.3× 10-2 mol H3PO4L×3 equivalentmol H3PO4=0.16 equivalentsLEquivalent mass = 13(molar mass of H3PO4) =13(97.09)=32.66 g$

Answer 27

(d)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(d)

Expert Solution

Answer to Problem 38E

The normality of $NaOH$ is $0.134N$

Explanation of Solution

Record the given data

Gram equivalent of $NaOH$ = $0.134$

Molarity of $NaOH$ = $0.134 M$

To calculate the normality of $NaOH$

$NaOH$ has mono acidic protons because it is a monoprotic acid Hence, its molarity can be one times

Therefore, the normality of $NaOH$ is given by,

$Normality = 0.134 mol NaOHL×1 equivalentmol NaOH=0.134 equivalentsLEquivalent mass = molar mass of NaOH 40 g$

Answer 28

(d)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

Answer 29

(d)

Expert Solution

Answer to Problem 38E

The normality of $NaOH$ is $0.134N$

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Record the given data

Gram equivalent of $NaOH$ = $0.134$

Molarity of $NaOH$ = $0.134 M$

To calculate the normality of $NaOH$

$NaOH$ has mono acidic protons because it is a monoprotic acid Hence, its molarity can be one times

Therefore, the normality of $NaOH$ is given by,

$Normality = 0.134 mol NaOHL×1 equivalentmol NaOH=0.134 equivalentsLEquivalent mass = molar mass of NaOH 40 g$

Answer 34

(e)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

(e)

Expert Solution

Answer to Problem 38E

The normality of $Ca(OH)2$ is $0.002605 N$

Explanation of Solution

Record the given data

Gram equivalent of $Ca(OH)2$ = $0.0104$

Molarity of $Ca(OH)2$ = $0.00521 M$

To calculate the normality of $Ca(OH)2$

$NaOH$ has di acidic protons because it is a diprotic acid Hence, its molarity can be two times

Therefore, the normality of $Ca(OH)2$ is given by,

$Normality = 0.00521 mol Ca(OH)2L×2 equivalentmol Ca(OH)2=0.0104 equivalentsLEquivalent mass = 12(molar mass of Ca(OH)2)=12(74.10)=37.05 g$

Answer 35

(e)

Interpretation Introduction

Interpretation:

The Normality and equivalent mass of given following solution has to be calculated.

Concept Introduction:

Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula,

$Normality(N)=Number of gram equivalent(in g)Volume of solution(in L)$

Answer 36

(e)

Expert Solution

Answer to Problem 38E

The normality of $Ca(OH)2$ is $0.002605 N$

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Record the given data

Gram equivalent of $Ca(OH)2$ = $0.0104$

Molarity of $Ca(OH)2$ = $0.00521 M$

To calculate the normality of $Ca(OH)2$

$NaOH$ has di acidic protons because it is a diprotic acid Hence, its molarity can be two times

Therefore, the normality of $Ca(OH)2$ is given by,

$Normality = 0.00521 mol Ca(OH)2L×2 equivalentmol Ca(OH)2=0.0104 equivalentsLEquivalent mass = 12(molar mass of Ca(OH)2)=12(74.10)=37.05 g$

Answer 41

Ch. 11 - Prob. 1RQ Ch. 11 - Using KF as an example, write equations that refer...Ch. 11 - Prob. 3RQ Ch. 11 - Prob. 4RQ Ch. 11 - Define the terms in Raoults law. Figure 10-9...Ch. 11 - In terms of Raoults law, distinguish between an...Ch. 11 - Vapor-pressure lowering is a colligative property,...Ch. 11 - What is osmotic pressure? How is osmotic pressure...Ch. 11 - Distinguish between a strong electrolyte, a weak...Ch. 11 - Prob. 10RQ

The Normality and equivalent mass of given following solution has to be calculated. Concept Introduction: Normality can be defined as number of gram equivalent to the volume of the solution. It can be given by formula, Normality(N)= Number of gram equivalent(in g) Volume of solution(in L)

Answer to Problem 38E

Explanation of Solution

Answer to Problem 38E

Explanation of Solution

Answer to Problem 38E

Explanation of Solution

Answer to Problem 38E

Explanation of Solution

Answer to Problem 38E

Explanation of Solution

Want to see more full solutions like this?

Chapter 11 Solutions