Chapter 9.5, Problem 9.8WE

(a)

Interpretation Introduction

Interpretation:

The molarity, volume, and number of moles of given glucose solution should be calculated.

Concept introduction:

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$\text{Molarity}\text{=}\frac{\text{No}\text{.mole}}{\text{volume}\text{(L)}}$

Mole:

Mole of solute is calculated by the ratio between mass of solute and molar mass of the solute.

$Mass=\frac{Mass}{Molarmass}$

Volume:

Volume of the solution is given by the ration between number of moles and molarity of solution.

$Volume=\frac{Mole}{Molarity}$

(b)

Interpretation Introduction

Interpretation:

The molarity, volume, and number of moles of given glucose solution should be calculated.

Concept introduction:

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$\text{Molarity}\text{=}\frac{\text{No}\text{.mole}}{\text{volume}\text{(L)}}$

Mole:

Mole of solute is calculated by the ratio between mass of solute and molar mass of the solute.

$Mass=\frac{Mass}{Molarmass}$

Volume:

Volume of the solution is given by the ration between number of moles and molarity of solution.

$Volume=\frac{Mole}{Molarity}$

(c)

Interpretation Introduction

Interpretation:

The molarity, volume, and number of moles of given glucose solution should be calculated.

Concept introduction:

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$\text{Molarity}\text{=}\frac{\text{No}\text{.mole}}{\text{volume}\text{(L)}}$

Mole:

Mole of solute is calculated by the ratio between mass of solute and molar mass of the solute.

$Mass=\frac{Mass}{Molarmass}$

Volume:

Volume of the solution is given by the ration between number of moles and molarity of solution.

$Volume=\frac{Mole}{Molarity}$