Number of grams of solute needed to prepare 250 m L o f 0.100 M N a B r solution has to be calculated. Concept Introduction: Molarity ( M ) can be calculated using following equation, M = m o l s o l u t e ( n ) L s o l u t i o n ( V L ) And number of mol can be calculated using given mass of the solute and molecular mass of the solute as follows, number of mol, n = mass of solute ( w ) molecular mass ( M ) On combining the above two equations another expression for molarity is obtained as below. M o l a r i t y = mass of solute m o l e c u l a r m a s s × v o l u m e ( L )

Question

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Answer 1

Question

Chapter 6, Problem 6.54QP

(a)

Interpretation Introduction

Interpretation:

Number of grams of solute needed to prepare 250 mL of 0.100 M NaBr solution has to be calculated.

Concept Introduction:

Molarity (M) can be calculated using following equation,

    M= mol solute(n)L solution(VL)

And number of mol can be calculated using given mass of the solute and molecular mass of the solute as follows,

    number of mol, n = mass of solute(w)molecular mass(M)

On combining the above two equations another expression for molarity is obtained as below.

    Molarity= mass of solutemolecular mass × volume(L)

(a)

Expert Solution

Answer to Problem 6.54QP

Number of grams of solute needed to prepare 250 mL of 0.100 M NaBr solution is 2.57 g.

Explanation of Solution

Molecular mass of sodium bromide is 102.894 g/mol.

Number of grams of solute needed to prepare 250 mL(0.250 L) of 0.100 M NaBr solution can be calculated as follows,

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.100 M) × (102.894 g/mol)×(0.250 L)= 2.57 g.

Amount solute required is 2.57 g.

(b)

Interpretation Introduction

Interpretation:

Number of grams of solute needed to prepare 250 mL of 0.200 M KOH solution has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 6.54QP

Number of grams of solute needed to prepare 250 mL of 0.200 M KOH solution is 2.8 g.

Explanation of Solution

Molecular mass of KOH is 56.1056 g/mol.

Number of grams of solute needed to prepare 250 mL of 0.200 M KOH solution can be calculated as follows,

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.200 M) × (56.1056 g/mol.)×(0.250 L)= 2.8 g.

Amount solute required is 2.8 g.

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Answer 2

Question

Chapter 6, Problem 6.54QP

(a)

Interpretation Introduction

Interpretation:

Number of grams of solute needed to prepare 250 mL of 0.100 M NaBr solution has to be calculated.

Concept Introduction:

Molarity (M) can be calculated using following equation,

    M= mol solute(n)L solution(VL)

And number of mol can be calculated using given mass of the solute and molecular mass of the solute as follows,

    number of mol, n = mass of solute(w)molecular mass(M)

On combining the above two equations another expression for molarity is obtained as below.

    Molarity= mass of solutemolecular mass × volume(L)

(a)

Expert Solution

Answer to Problem 6.54QP

Number of grams of solute needed to prepare 250 mL of 0.100 M NaBr solution is 2.57 g.

Explanation of Solution

Molecular mass of sodium bromide is 102.894 g/mol.

Number of grams of solute needed to prepare 250 mL(0.250 L) of 0.100 M NaBr solution can be calculated as follows,

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.100 M) × (102.894 g/mol)×(0.250 L)= 2.57 g.

Amount solute required is 2.57 g.

(b)

Interpretation Introduction

Interpretation:

Number of grams of solute needed to prepare 250 mL of 0.200 M KOH solution has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 6.54QP

Number of grams of solute needed to prepare 250 mL of 0.200 M KOH solution is 2.8 g.

Explanation of Solution

Molecular mass of KOH is 56.1056 g/mol.

Number of grams of solute needed to prepare 250 mL of 0.200 M KOH solution can be calculated as follows,

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.200 M) × (56.1056 g/mol.)×(0.250 L)= 2.8 g.

Amount solute required is 2.8 g.

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Answer 3

Question

Chapter 6, Problem 6.54QP

(a)

Interpretation Introduction

Interpretation:

Number of grams of solute needed to prepare 250 mL of 0.100 M NaBr solution has to be calculated.

Concept Introduction:

Molarity (M) can be calculated using following equation,

    M= mol solute(n)L solution(VL)

And number of mol can be calculated using given mass of the solute and molecular mass of the solute as follows,

    number of mol, n = mass of solute(w)molecular mass(M)

On combining the above two equations another expression for molarity is obtained as below.

    Molarity= mass of solutemolecular mass × volume(L)

(a)

Expert Solution

Answer to Problem 6.54QP

Number of grams of solute needed to prepare 250 mL of 0.100 M NaBr solution is 2.57 g.

Explanation of Solution

Molecular mass of sodium bromide is 102.894 g/mol.

Number of grams of solute needed to prepare 250 mL(0.250 L) of 0.100 M NaBr solution can be calculated as follows,

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.100 M) × (102.894 g/mol)×(0.250 L)= 2.57 g.

Amount solute required is 2.57 g.

(b)

Interpretation Introduction

Interpretation:

Number of grams of solute needed to prepare 250 mL of 0.200 M KOH solution has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 6.54QP

Number of grams of solute needed to prepare 250 mL of 0.200 M KOH solution is 2.8 g.

Explanation of Solution

Molecular mass of KOH is 56.1056 g/mol.

Number of grams of solute needed to prepare 250 mL of 0.200 M KOH solution can be calculated as follows,

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.200 M) × (56.1056 g/mol.)×(0.250 L)= 2.8 g.

Amount solute required is 2.8 g.

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Answer 4

Question

Chapter 6, Problem 6.54QP

(a)

Interpretation Introduction

Interpretation:

Number of grams of solute needed to prepare 250 mL of 0.100 M NaBr solution has to be calculated.

Concept Introduction:

Molarity (M) can be calculated using following equation,

    M= mol solute(n)L solution(VL)

And number of mol can be calculated using given mass of the solute and molecular mass of the solute as follows,

    number of mol, n = mass of solute(w)molecular mass(M)

On combining the above two equations another expression for molarity is obtained as below.

    Molarity= mass of solutemolecular mass × volume(L)

(a)

Expert Solution

Answer to Problem 6.54QP

Number of grams of solute needed to prepare 250 mL of 0.100 M NaBr solution is 2.57 g.

Explanation of Solution

Molecular mass of sodium bromide is 102.894 g/mol.

Number of grams of solute needed to prepare 250 mL(0.250 L) of 0.100 M NaBr solution can be calculated as follows,

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.100 M) × (102.894 g/mol)×(0.250 L)= 2.57 g.

Amount solute required is 2.57 g.

(b)

Interpretation Introduction

Interpretation:

Number of grams of solute needed to prepare 250 mL of 0.200 M KOH solution has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 6.54QP

Number of grams of solute needed to prepare 250 mL of 0.200 M KOH solution is 2.8 g.

Explanation of Solution

Molecular mass of KOH is 56.1056 g/mol.

Number of grams of solute needed to prepare 250 mL of 0.200 M KOH solution can be calculated as follows,

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.200 M) × (56.1056 g/mol.)×(0.250 L)= 2.8 g.

Amount solute required is 2.8 g.

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Answer 5

Chapter 6, Problem 6.54QP

(a)

Interpretation Introduction

Interpretation:

Number of grams of solute needed to prepare 250 mL of 0.100 M NaBr solution has to be calculated.

Concept Introduction:

Molarity (M) can be calculated using following equation,

    M= mol solute(n)L solution(VL)

And number of mol can be calculated using given mass of the solute and molecular mass of the solute as follows,

    number of mol, n = mass of solute(w)molecular mass(M)

On combining the above two equations another expression for molarity is obtained as below.

    Molarity= mass of solutemolecular mass × volume(L)

(a)

Expert Solution

Answer to Problem 6.54QP

Number of grams of solute needed to prepare 250 mL of 0.100 M NaBr solution is 2.57 g.

Explanation of Solution

Molecular mass of sodium bromide is 102.894 g/mol.

Number of grams of solute needed to prepare 250 mL(0.250 L) of 0.100 M NaBr solution can be calculated as follows,

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.100 M) × (102.894 g/mol)×(0.250 L)= 2.57 g.

Amount solute required is 2.57 g.

(b)

Interpretation Introduction

Interpretation:

Number of grams of solute needed to prepare 250 mL of 0.200 M KOH solution has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 6.54QP

Number of grams of solute needed to prepare 250 mL of 0.200 M KOH solution is 2.8 g.

Explanation of Solution

Molecular mass of KOH is 56.1056 g/mol.

Number of grams of solute needed to prepare 250 mL of 0.200 M KOH solution can be calculated as follows,

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.200 M) × (56.1056 g/mol.)×(0.250 L)= 2.8 g.

Amount solute required is 2.8 g.

Answer 6

Chapter 6, Problem 6.54QP

(a)

Interpretation Introduction

Interpretation:

Number of grams of solute needed to prepare 250 mL of 0.100 M NaBr solution has to be calculated.

Concept Introduction:

Molarity (M) can be calculated using following equation,

    M= mol solute(n)L solution(VL)

And number of mol can be calculated using given mass of the solute and molecular mass of the solute as follows,

    number of mol, n = mass of solute(w)molecular mass(M)

On combining the above two equations another expression for molarity is obtained as below.

    Molarity= mass of solutemolecular mass × volume(L)

(a)

Expert Solution

Answer to Problem 6.54QP

Number of grams of solute needed to prepare 250 mL of 0.100 M NaBr solution is 2.57 g.

Explanation of Solution

Molecular mass of sodium bromide is 102.894 g/mol.

Number of grams of solute needed to prepare 250 mL(0.250 L) of 0.100 M NaBr solution can be calculated as follows,

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.100 M) × (102.894 g/mol)×(0.250 L)= 2.57 g.

Amount solute required is 2.57 g.

Answer 7

Chapter 6, Problem 6.54QP

(a)

Interpretation Introduction

Interpretation:

Number of grams of solute needed to prepare 250 mL of 0.100 M NaBr solution has to be calculated.

Concept Introduction:

Molarity (M) can be calculated using following equation,

    M= mol solute(n)L solution(VL)

And number of mol can be calculated using given mass of the solute and molecular mass of the solute as follows,

    number of mol, n = mass of solute(w)molecular mass(M)

On combining the above two equations another expression for molarity is obtained as below.

    Molarity= mass of solutemolecular mass × volume(L)

Answer 8

(a)

Expert Solution

Answer to Problem 6.54QP

Number of grams of solute needed to prepare 250 mL of 0.100 M NaBr solution is 2.57 g.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Molecular mass of sodium bromide is 102.894 g/mol.

Number of grams of solute needed to prepare 250 mL(0.250 L) of 0.100 M NaBr solution can be calculated as follows,

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.100 M) × (102.894 g/mol)×(0.250 L)= 2.57 g.

Amount solute required is 2.57 g.

Answer 13

(b)

Interpretation Introduction

Interpretation:

Number of grams of solute needed to prepare 250 mL of 0.200 M KOH solution has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 6.54QP

Number of grams of solute needed to prepare 250 mL of 0.200 M KOH solution is 2.8 g.

Explanation of Solution

Molecular mass of KOH is 56.1056 g/mol.

Number of grams of solute needed to prepare 250 mL of 0.200 M KOH solution can be calculated as follows,

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.200 M) × (56.1056 g/mol.)×(0.250 L)= 2.8 g.

Amount solute required is 2.8 g.

Answer 14

(b)

Interpretation Introduction

Interpretation:

Number of grams of solute needed to prepare 250 mL of 0.200 M KOH solution has to be calculated.

Concept Introduction:

Refer to part (a).

Answer 15

(b)

Expert Solution

Answer to Problem 6.54QP

Number of grams of solute needed to prepare 250 mL of 0.200 M KOH solution is 2.8 g.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Molecular mass of KOH is 56.1056 g/mol.

Number of grams of solute needed to prepare 250 mL of 0.200 M KOH solution can be calculated as follows,

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.200 M) × (56.1056 g/mol.)×(0.250 L)= 2.8 g.