(a) Interpretation: The volume of 5 M sucrose solution required to prepare 45 mL of 4 M solution is to be calculated. Concept introduction: When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows: Number of moles = Molarity × Volume This can be rewritten as follows: MV = constant Thus the dilution formula is given as follows: M 1 V 1 = M 2 V 2

Question

Want to see the full answer?

Answer 1

Question

Chapter 8, Problem 8.92P

Interpretation Introduction

(a)

Interpretation:

The volume of 5 M sucrose solution required to prepare 45 mL of 4 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Interpretation Introduction

(b)

Interpretation:

The volume of 5 M sucrose solution required to prepare 150 mL of 0.5 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Interpretation Introduction

(c)

Interpretation:

The volume of 0.025 M solution required to prepare 1.2 L of 0.025 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Interpretation Introduction

(d)

Interpretation:

The volume of 5 M sucrose solution required to prepare 750 mL of 1 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Curved arrows are used to illustrate the flow of electrons. Using the provided structures, draw the curved arrows that epict the mechanistic steps for the proton transfer between a hydronium ion and a pi bond. Draw any missing organic structures in the empty boxes. Be sure to account for all lone-pairs and charges as well as bond-breaking and bond-making steps. 2 56°F Mostly cloudy F1 Drawing Arrows > Q Search F2 F3 F4 ▷11 H. H : CI: H + Undo Reset Done DELL

Calculate the chemical shifts in 13C and 1H NMR for 4-chloropropiophenone ? Write structure and label hydrogens and carbons. Draw out the benzene ring structure when doing it

1) Calculate the longest and shortest wavelengths in the Lyman and Paschen series. 2) Calculate the ionization energy of He* and L2+ ions in their ground states. 3) Calculate the kinetic energy of the electron emitted upon irradiation of a H-atom in ground state by a 50-nm radiation.

Answer 2

Question

Chapter 8, Problem 8.92P

Interpretation Introduction

(a)

Interpretation:

The volume of 5 M sucrose solution required to prepare 45 mL of 4 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Interpretation Introduction

(b)

Interpretation:

The volume of 5 M sucrose solution required to prepare 150 mL of 0.5 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Interpretation Introduction

(c)

Interpretation:

The volume of 0.025 M solution required to prepare 1.2 L of 0.025 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Interpretation Introduction

(d)

Interpretation:

The volume of 5 M sucrose solution required to prepare 750 mL of 1 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 8, Problem 8.92P

Interpretation Introduction

(a)

Interpretation:

The volume of 5 M sucrose solution required to prepare 45 mL of 4 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Interpretation Introduction

(b)

Interpretation:

The volume of 5 M sucrose solution required to prepare 150 mL of 0.5 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Interpretation Introduction

(c)

Interpretation:

The volume of 0.025 M solution required to prepare 1.2 L of 0.025 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Interpretation Introduction

(d)

Interpretation:

The volume of 5 M sucrose solution required to prepare 750 mL of 1 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 8, Problem 8.92P

Interpretation Introduction

(a)

Interpretation:

The volume of 5 M sucrose solution required to prepare 45 mL of 4 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Interpretation Introduction

(b)

Interpretation:

The volume of 5 M sucrose solution required to prepare 150 mL of 0.5 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Interpretation Introduction

(c)

Interpretation:

The volume of 0.025 M solution required to prepare 1.2 L of 0.025 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Interpretation Introduction

(d)

Interpretation:

The volume of 5 M sucrose solution required to prepare 750 mL of 1 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Chapter 8, Problem 8.92P

Interpretation Introduction

(a)

Interpretation:

The volume of 5 M sucrose solution required to prepare 45 mL of 4 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Interpretation Introduction

(b)

Interpretation:

The volume of 5 M sucrose solution required to prepare 150 mL of 0.5 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Interpretation Introduction

(c)

Interpretation:

The volume of 0.025 M solution required to prepare 1.2 L of 0.025 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Interpretation Introduction

(d)

Interpretation:

The volume of 5 M sucrose solution required to prepare 750 mL of 1 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Answer 6

Chapter 8, Problem 8.92P

Interpretation Introduction

(a)

Interpretation:

The volume of 5 M sucrose solution required to prepare 45 mL of 4 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Answer 7

Chapter 8, Problem 8.92P

Interpretation Introduction

(a)

Interpretation:

The volume of 5 M sucrose solution required to prepare 45 mL of 4 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Answer 8

Interpretation Introduction

(b)

Interpretation:

The volume of 5 M sucrose solution required to prepare 150 mL of 0.5 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Answer 9

Interpretation Introduction

(b)

Interpretation:

The volume of 5 M sucrose solution required to prepare 150 mL of 0.5 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Answer 10

Interpretation Introduction

(c)

Interpretation:

The volume of 0.025 M solution required to prepare 1.2 L of 0.025 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Answer 11

Interpretation Introduction

(c)

Interpretation:

The volume of 0.025 M solution required to prepare 1.2 L of 0.025 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Answer 12

Interpretation Introduction

(d)

Interpretation:

The volume of 5 M sucrose solution required to prepare 750 mL of 1 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Answer 13

Interpretation Introduction

(d)

Interpretation:

The volume of 5 M sucrose solution required to prepare 750 mL of 1 M solution is to be calculated.

Concept introduction:

When a solution is diluted the number of moles of the solute remains constant. Mathematically, this can be expressed as follows:

Number of moles=Molarity×Volume

This can be rewritten as follows:

MV=constant

Thus the dilution formula is given as follows:

M1V1=M2V2

Answer 14

Expert Solution & Answer

Answer 15

Expert Solution & Answer

Answer 16

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 17

Ch. 8.1 - Classify each substance as a heterogeneous...Ch. 8.1 - Classify each product as a solution, colloid, or...Ch. 8.2 - Consider the following diagrams for an aqueous...Ch. 8.2 - Classify each solution as an electrolyte or...Ch. 8.2 - Using the given number of moles, determine how...Ch. 8.2 - Prob. 8.6P Ch. 8.2 - A solution contains the following ions:...Ch. 8.2 - If a solution contains 125 mEq of Na+ per liter,...Ch. 8.3 - Which compounds are water soluble? a. NaNO3 b. CH4...Ch. 8.3 - Prob. 8.10P

Solution Summary: The author explains that when a solution is diluted the number of moles of the solute remains constant.

Want to see the full answer?

Chapter 8 Solutions