Number of grams of solute needed to prepare 250 m L o f 0.100 M N a C l 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 6.53QP

(a)

Interpretation Introduction

Interpretation:

Number of grams of solute needed to prepare 250 mL of 0.100 M NaCl 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 6.53QP

Number of grams of solute needed to prepare 250 mL of 0.100 M NaCl solution is 1.461 g.

Explanation of Solution

Molecular mass of sodium chloride is 58.44 g.

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

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.100 M) × (58.44g/mol)×(0.250 L)= 1.461 g.

Amount solute required is 1.461 g.

(b)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 6.53QP

Number of grams of solute needed to prepare 250 mL of 0.200 M glucose solution is 9.0078 g.

Explanation of Solution

Molecular mass of glucose is 180.156 g/mol.

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

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.200 M) × (180.156 g/mol)×(0.250 L)= 9.0078 g.

Amount solute required is 9.0078 g.

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Answer 2

Question

Chapter 6, Problem 6.53QP

(a)

Interpretation Introduction

Interpretation:

Number of grams of solute needed to prepare 250 mL of 0.100 M NaCl 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 6.53QP

Number of grams of solute needed to prepare 250 mL of 0.100 M NaCl solution is 1.461 g.

Explanation of Solution

Molecular mass of sodium chloride is 58.44 g.

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

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.100 M) × (58.44g/mol)×(0.250 L)= 1.461 g.

Amount solute required is 1.461 g.

(b)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 6.53QP

Number of grams of solute needed to prepare 250 mL of 0.200 M glucose solution is 9.0078 g.

Explanation of Solution

Molecular mass of glucose is 180.156 g/mol.

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

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.200 M) × (180.156 g/mol)×(0.250 L)= 9.0078 g.

Amount solute required is 9.0078 g.

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Answer 3

Question

Chapter 6, Problem 6.53QP

(a)

Interpretation Introduction

Interpretation:

Number of grams of solute needed to prepare 250 mL of 0.100 M NaCl 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 6.53QP

Number of grams of solute needed to prepare 250 mL of 0.100 M NaCl solution is 1.461 g.

Explanation of Solution

Molecular mass of sodium chloride is 58.44 g.

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

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.100 M) × (58.44g/mol)×(0.250 L)= 1.461 g.

Amount solute required is 1.461 g.

(b)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 6.53QP

Number of grams of solute needed to prepare 250 mL of 0.200 M glucose solution is 9.0078 g.

Explanation of Solution

Molecular mass of glucose is 180.156 g/mol.

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

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.200 M) × (180.156 g/mol)×(0.250 L)= 9.0078 g.

Amount solute required is 9.0078 g.

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Answer 4

Question

Chapter 6, Problem 6.53QP

(a)

Interpretation Introduction

Interpretation:

Number of grams of solute needed to prepare 250 mL of 0.100 M NaCl 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 6.53QP

Number of grams of solute needed to prepare 250 mL of 0.100 M NaCl solution is 1.461 g.

Explanation of Solution

Molecular mass of sodium chloride is 58.44 g.

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

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.100 M) × (58.44g/mol)×(0.250 L)= 1.461 g.

Amount solute required is 1.461 g.

(b)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 6.53QP

Number of grams of solute needed to prepare 250 mL of 0.200 M glucose solution is 9.0078 g.

Explanation of Solution

Molecular mass of glucose is 180.156 g/mol.

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

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.200 M) × (180.156 g/mol)×(0.250 L)= 9.0078 g.

Amount solute required is 9.0078 g.

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Answer 5

Chapter 6, Problem 6.53QP

(a)

Interpretation Introduction

Interpretation:

Number of grams of solute needed to prepare 250 mL of 0.100 M NaCl 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 6.53QP

Number of grams of solute needed to prepare 250 mL of 0.100 M NaCl solution is 1.461 g.

Explanation of Solution

Molecular mass of sodium chloride is 58.44 g.

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

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.100 M) × (58.44g/mol)×(0.250 L)= 1.461 g.

Amount solute required is 1.461 g.

(b)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 6.53QP

Number of grams of solute needed to prepare 250 mL of 0.200 M glucose solution is 9.0078 g.

Explanation of Solution

Molecular mass of glucose is 180.156 g/mol.

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

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.200 M) × (180.156 g/mol)×(0.250 L)= 9.0078 g.

Amount solute required is 9.0078 g.

Answer 6

Chapter 6, Problem 6.53QP

(a)

Interpretation Introduction

Interpretation:

Number of grams of solute needed to prepare 250 mL of 0.100 M NaCl 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 6.53QP

Number of grams of solute needed to prepare 250 mL of 0.100 M NaCl solution is 1.461 g.

Explanation of Solution

Molecular mass of sodium chloride is 58.44 g.

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

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.100 M) × (58.44g/mol)×(0.250 L)= 1.461 g.

Amount solute required is 1.461 g.

Answer 7

Chapter 6, Problem 6.53QP

(a)

Interpretation Introduction

Interpretation:

Number of grams of solute needed to prepare 250 mL of 0.100 M NaCl 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 6.53QP

Number of grams of solute needed to prepare 250 mL of 0.100 M NaCl solution is 1.461 g.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Molecular mass of sodium chloride is 58.44 g.

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

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.100 M) × (58.44g/mol)×(0.250 L)= 1.461 g.

Amount solute required is 1.461 g.

Answer 13

(b)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 6.53QP

Number of grams of solute needed to prepare 250 mL of 0.200 M glucose solution is 9.0078 g.

Explanation of Solution

Molecular mass of glucose is 180.156 g/mol.

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

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.200 M) × (180.156 g/mol)×(0.250 L)= 9.0078 g.

Amount solute required is 9.0078 g.

Answer 14

(b)

Interpretation Introduction

Interpretation:

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

Concept Introduction:

Refer to part (a).

Answer 15

(b)

Expert Solution

Answer to Problem 6.53QP

Number of grams of solute needed to prepare 250 mL of 0.200 M glucose solution is 9.0078 g.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Molecular mass of glucose is 180.156 g/mol.

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

Molarity= mass of solutemolecular mass × volume(L) mass of solute = (Molarity)×(molecular mass) × (volume(L))= (0.200 M) × (180.156 g/mol)×(0.250 L)= 9.0078 g.