Chapter 7, Problem 7.31E

Interpretation Introduction

(a)

Interpretation:

The concentration in $\%\left(\text{w}/\text{w}\right)$ of $5.20\text{g}$ of ${\text{CaCl}}_{\text{2}}$ dissolved in $125\text{mL}$ of water is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume $\left(\text{\%}\text{w}/\text{v}\right)$, percent volume by volume $\left(\text{\%}\text{v}/\text{v}\right)$ and molarity $\left(\text{M}\right)$. When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight $\left(\text{\%}\text{w}/\text{w}\right)$.

Answer to Problem 7.31E

The concentration in $\%\left(\text{w}/\text{w}\right)$ of $5.20\text{g}$ of ${\text{CaCl}}_{\text{2}}$ dissolved in $125\text{mL}$ of water is $3.99\%\text{w}/\text{w}$.

Explanation of Solution

The formula to calculate concentration in $\%\left(\text{w}/\text{w}\right)$ is given below as,

$\%\left(\text{w}/\text{w}\right)=\frac{\text{Solute}\text{Mass}}{\text{Solution}\text{Mass}}\times 100$

The density of water is $1.00\text{g}/\text{mL}$. Thus, the mass of water is equal to its volume. Hence in this case the mass of solution is taken as $125\text{g}$. Substitute the value of mass of salt and mass of solution in the above equation as follows.

$\begin{array}{rll}\%\left(\text{w}/\text{w}\right)& =& \frac{\text{Mass}{\text{of CaCl}}_{\text{2}}}{\text{Solution}\text{Mass}}\times 100\\ & =& \frac{5.20\text{g}}{\left(125+5.20\right)\text{g}}\times 100\\ & =& \frac{5.20\text{g}}{130.2\text{g}}\times 100\\ & =& 3.99\%\text{w}/\text{w}\end{array}$

Conclusion

The concentration in $\%\left(\text{w}/\text{w}\right)$ of $5.20\text{g}$ of ${\text{CaCl}}_{\text{2}}$ dissolved in $125\text{mL}$ of water is $3.99\%\text{w}/\text{w}$.

Interpretation Introduction

(b)

Interpretation:

The concentration in $\%\left(\text{w}/\text{w}\right)$ of $0.200\text{mol}$ of solid $\text{KBr}$ dissolved in $200.\text{mL}$ of water is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume $\left(\text{\%}\text{w}/\text{v}\right)$, percent volume by volume $\left(\text{\%}\text{v}/\text{v}\right)$ and molarity $\left(\text{M}\right)$. When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight $\left(\text{\%}\text{w}/\text{w}\right)$.

Answer to Problem 7.31E

The concentration in $\%\left(\text{w}/\text{w}\right)$ of $0.200\text{mol}$ of solid $\text{KBr}$ dissolved in $200.\text{mL}$ of water is $10.63\%\text{w}/\text{w}$.

Explanation of Solution

The formula to calculate number of moles of solutes is given below as,

$\text{Moles}\text{of}\text{solute}=\frac{\text{Mass}\text{of}\text{solute}}{\text{Molar}\text{mass}\text{of}\text{solute}}$

Rearrange the above equation for mass of solute as follows.

$\begin{array}{l}\text{Moles}\text{of}\text{solute}=\frac{\text{Mass}\text{of}\text{solute}}{\text{Molar}\text{mass}\text{of}\text{solute}}\\ \text{Mass}\text{of}\text{solute}=\text{Moles}\text{of}\text{solute}\times \text{Molar}\text{mass}\text{of}\text{solute}\end{array}$

The molar mass of $\text{KBr}$ can be calculated as follows.

$\begin{array}{rll}\text{KBr}& =& \left(1\times \text{K}\right)+\left(1\times \text{Br}\right)\\ & =& \left(1\times 39.0983\right)+\left(79.904\times 1\right)\\ & =& 119.002\text{g}/\text{mol}\end{array}$

Substitute the values in the above equation as follows.

$\begin{array}{rll}\text{Mass}\text{of}\text{KBr}& =& \text{Moles}\text{of}\text{KBr}\times \text{Molar}\text{mass}\text{of}\text{KBr}\\ & =& 0.200\text{mol}\times 119.002\text{g}/\text{mol}\\ & =& 23.8\text{g}\end{array}$

The formula to calculate concentration in $\%\left(\text{w}/\text{w}\right)$ is given below as,

$\%\left(\text{w}/\text{w}\right)=\frac{\text{Solute}\text{Mass}}{\text{Solution}\text{Mass}}\times 100$

The density of water is $1.00\text{g}/\text{mL}$. Thus, the mass of water is equal to its volume. Hence in this case the mass of solution is taken as $200.\text{g}$. Substitute the value of mass of solute and mass of solution in the above equation as follows.

$\begin{array}{rll}\%\left(\text{w}/\text{w}\right)& =& \frac{\text{Mass}\text{of}\text{KBr}}{\text{Solution}\text{Mass}}\times 100\\ & =& \frac{23.8\text{g}}{\left(200+23.8\right)\text{g}}\times 100\\ & =& \frac{23.8\text{g}}{223.8\text{g}}\times 100\\ & =& 10.63\%\text{w}/\text{w}\end{array}$

Conclusion

The concentration in $\%\left(\text{w}/\text{w}\right)$ of $0.200\text{mol}$ of solid $\text{KBr}$ dissolved in $200.\text{mL}$ of water is $10.63\%\text{w}/\text{w}$.

Interpretation Introduction

(c)

Interpretation:

The concentration in $\%\left(\text{w}/\text{w}\right)$ of $50.0\text{g}$ of solid dissolved in $250.\text{mL}$ of water is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume $\left(\text{\%}\text{w}/\text{v}\right)$, percent volume by volume $\left(\text{\%}\text{v}/\text{v}\right)$ and molarity $\left(\text{M}\right)$. When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight $\left(\text{\%}\text{w}/\text{w}\right)$.

Answer to Problem 7.31E

The concentration in $\%\left(\text{w}/\text{w}\right)$ of $50.0\text{g}$ of solid dissolved in $250.\text{mL}$ of water is $16.67\%\text{w}/\text{w}$.

Explanation of Solution

The formula to calculate concentration in $\%\left(\text{w}/\text{w}\right)$ is given below as,

$\%\left(\text{w}/\text{w}\right)=\frac{\text{Solute}\text{Mass}}{\text{Solution}\text{Mass}}\times 100$

The density of water is $1.00\text{g}/\text{mL}$. Thus, the mass of water is equal to its volume. Hence in this case the mass of solution is taken as $250\text{g}$. Substitute the value of mass of solute and mass of solution in the above equation as follows.

$\begin{array}{rll}\%\left(\text{w}/\text{w}\right)& =& \frac{\text{Solute}\text{Mass}}{\text{Solution}\text{Mass}}\times 100\\ & =& \frac{50.0\text{g}}{\left(250+50\right)\text{g}}\times 100\\ & =& \frac{50\text{g}}{300\text{g}}\times 100\\ & =& 16.67\%\text{w}/\text{w}\end{array}$

Conclusion

The concentration in $\%\left(\text{w}/\text{w}\right)$ of $50.0\text{g}$ of solid dissolved in $250.\text{mL}$ of water is $16.67\%\text{w}/\text{w}$.

Interpretation Introduction

(d)

Interpretation:

The concentration in $\%\left(\text{w}/\text{w}\right)$ of $10.0\text{mL}$ of ethyl alcohol (density $=0.789\text{g}/\text{mL}$) mixed with $10.0\text{mL}$ of ethylene glycol (density $=1.11\text{g}/\text{mL}$) is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, it is expressed in terms of percent weight by volume $\left(\text{\%}\text{w}/\text{v}\right)$, percent volume by volume $\left(\text{\%}\text{v}/\text{v}\right)$ and molarity $\left(\text{M}\right)$. When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight $\left(\text{\%}\text{w}/\text{w}\right)$.

Answer to Problem 7.31E

The concentration in $\%\left(\text{w}/\text{w}\right)$ of $10.0\text{mL}$ of ethyl alcohol (density $=0.789\text{g}/\text{mL}$) mixed with $10.0\text{mL}$ of ethylene glycol (density $=1.11\text{g}/\text{mL}$) is $41.55\%\text{w}/\text{w}$.

Explanation of Solution

The formula to calculate concentration in $\%\left(\text{w}/\text{w}\right)$ is given below as,

$\%\left(\text{w}/\text{w}\right)=\frac{\text{Solute}\text{Mass}}{\text{Solution}\text{Mass}}\times 100$ …(1)

The formula to calculate density is given below as,

$d=\frac{m}{V}$

Where,

• $m$ is the mass of the substance.

• $V$ is the volume of the substance.

• $d$ is the density of the substance.

The density of ethyl alcohol is $0.789\text{g}/\text{mL}$. Thus, the mass of $10.0\text{mL}$ of ethyl alcohol can be calculated as follows.

$\begin{array}{rll}d& =& \frac{m}{V}\\ m& =& d\times V\\ & =& 0.789\text{g}/\text{mL}\times 10.0\text{mL}\\ & =& 7.89\text{g}\end{array}$

The density of ethylene glycol is $1.11\text{g}/\text{mL}$. Thus, the mass of $10.0\text{mL}$ of ethylene glycol can be calculated as follows.

$\begin{array}{rll}d& =& \frac{m}{V}\\ m& =& d\times V\\ & =& 1.11\text{g}/\text{mL}\times 10.0\text{mL}\\ & =& 11.1\text{g}\end{array}$

Substitute the value of mass of ethyl alcohol and mass of solution in the equation (1) as follows.

$\begin{array}{rll}\%\left(\text{w}/\text{w}\right)& =& \frac{\text{Mass}\text{of}\text{ethyl alcohol}}{\text{Solution}\text{Mass}}\times 100\\ & =& \frac{7.89\text{g}}{\left(11.1\text{g}+7.89\right)\text{g}}\times 100\\ & =& \frac{7.89\text{g}}{18.99\text{g}}\times 100\\ & =& 41.55\%\text{w}/\text{w}\end{array}$

Conclusion

The concentration in $\%\left(\text{w}/\text{w}\right)$ of $10.0\text{mL}$ of ethyl alcohol (density $=0.789\text{g}/\text{mL}$) mixed with $10.0\text{mL}$ of ethylene glycol (density $=1.11\text{g}/\text{mL}$) is $41.55\%\text{w}/\text{w}$.