Chapter 9.5, Problem 8PPA

(a)

Interpretation Introduction

Interpretation:

The molarity, volume, and number of moles of given sucrose 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.

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

Answer to Problem 8PPA

Molarity of sucrose solution is $0.1374M$.

Explanation of Solution

The molarity of $5.00L$solution contains $235g$ of sucrose,

Molar mass of the sucrose is $\text{342}\text{.3}\text{g}$

$\begin{array}{l}Moleofthesucroseintakengram =\frac{235.0g}{342.3g}\\ =0.687moles\end{array}$

Molarity of 5.00 L solution contains 235 g of sucrose is,

$\begin{array}{l}=\frac{0.687mole}{5.00L}\\ =0.1374M\end{array}$

The molar mass and taken mass of sucrose is plugged in the above equation to gives mole of sucrose present in taken gram.

The calculated mole of sucrose is divided by volume of the solution to gives molarity of sucrose solution.

Molarity of sucrose solution is $0.1374M$.

(b)

Interpretation Introduction

Interpretation:

The molarity, volume, and number of moles of given sucrose 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.

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

Answer to Problem 8PPA

Volume of the sucrose solution is $9.2L$

Explanation of Solution

The volume of solution contains $\text{1}\text{.26 mole}$ of sucrose,

$\begin{array}{l}Volume=\frac{1.026mole}{0.374M}\\ =9.2L\end{array}$

The given mole is divided by calculated molarity to gives volume of the sucrose solution.

Volume of the sucrose solution is $9.2L$

(c)

Interpretation Introduction

Interpretation:

The molarity, volume, and number of moles of given sucrose 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.

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

Answer to Problem 8PPA

Mole of the sucrose solution is0.26 mole

Explanation of Solution

The mole of $1.89L$solution of sucrose,

$\begin{array}{l}=1.89L\times 0.1374M\\ =0.26mole\end{array}$

The calculated molarity is multiplied with volume to give mole of the sucrose solution.

 Mole of the sucrose solution is $0.26mole$.