Chapter 13, Problem 56GQ

(a)

Interpretation Introduction

Interpretation: The weight percentage of $NaCl$ has to be determined.

Concept introduction:

Mole fraction: Amount of that component divided by the total amount of all of the components of the mixture $\left({\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}\right)$.

  ${\text{Mole fraction of A (χ}}_{\text{A}}\text{)= }\frac{{\text{n}}_{\text{A}}}{{\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}}$

Weight percent: The mass of one component divided by the total mass of the mixture, multiplied by $100\%$

  $\text{weight \% A = }\frac{\text{Mass of A}}{\text{ Mass of A + Mass of B + Mass of C + }\mathrm{...}}\text{×100\%}$

The number of moles of any substance can be determined using the equation

  $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent.

  $\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

Answer to Problem 56GQ

Weight percentage of $NaCl$ is $\text{13}\text{.}\text{5}\text{\%}$

Explanation of Solution

Given,

  Given mass of $NaCl$ is $1130g$

  Given mass of ice is $7250g=7.25kg$

Weight percentage of $NaCl$ is calculated,

  $\begin{array}{c}\text{Weight percentage}\text{of}\text{NaCl}\text{= }\frac{\text{Mass of NaCl}}{\text{ Mass of NaCl + Mass of ice }}\text{×100\%}\\ \text{=}\frac{\text{1130}\text{g}}{\left(\text{1130}\text{g}\text{+}\text{7250}\text{g}\right)}\text{×100\%}\\ \text{=}\text{13}\text{.}\text{5}\text{\%}\end{array}$

Weight percentage of $NaCl$ is $\text{13}\text{.}\text{5}\text{\%}$

(b)

Interpretation Introduction

Interpretation: The mole fraction of $NaCl$ has to be determined.

Concept introduction:

Mole fraction: Amount of that component divided by the total amount of all of the components of the mixture $\left({\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}\right)$.

  ${\text{Mole fraction of A (χ}}_{\text{A}}\text{)= }\frac{{\text{n}}_{\text{A}}}{{\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}}$

Weight percent: The mass of one component divided by the total mass of the mixture, multiplied by $100\%$

  $\text{weight \% A = }\frac{\text{Mass of A}}{\text{ Mass of A + Mass of B + Mass of C + }\mathrm{...}}\text{×100\%}$

The number of moles of any substance can be determined using the equation

  $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent.

  $\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

Answer to Problem 56GQ

Mole fraction of $NaCl$ is $\text{0}\text{.046}$

Explanation of Solution

Given,

  Given mass of $NaCl$ is $1130g$

  Given mass of ice is $7250g=7.25kg$

  Molar mass of $NaCl$ is $\text{58}\text{.44}\text{g/mol}$

  Molar mass of ice is $\text{18}\text{.02}\text{g/mol}$

Number of moles of $NaCl$ is,

  $\begin{array}{c}{\text{n}}_{NaCl}\text{=}\frac{\text{1130}\text{g}}{\text{58}\text{.44}\text{g/mol}}\\ =19.336mol\end{array}$

Number of moles of ice is,

  $\begin{array}{c}{\text{n}}_{ice}\text{=}\frac{\text{7250}\text{g}}{\text{18}\text{.02}\text{g/mol}}\\ =402.33mol\end{array}$

Mole fraction of $NaCl$ is calculated,

$\begin{array}{c}{\text{Mole fraction of NaCl (χ}}_{\text{NaCl}}\text{)= }\frac{{\text{n}}_{\text{NaCl}}}{{\text{ n}}_{\text{NaCl}}{\text{ + n}}_{\text{ice}}}\\ \text{=}\frac{\text{19}\text{.336}\text{mol}}{\left(\text{19}\text{.336}\text{mol}\text{+}\text{402}\text{.33}\text{mol}\right)}\\ \text{=}\text{0}\text{.046}\end{array}$

Mole fraction of $NaCl$ is $\text{0}\text{.046}$

(c)

Interpretation Introduction

Interpretation: The molality of the solution of $NaCl$ has to be determined.

Concept introduction:

Mole fraction: Amount of that component divided by the total amount of all of the components of the mixture $\left({\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}\right)$.

  ${\text{Mole fraction of A (χ}}_{\text{A}}\text{)= }\frac{{\text{n}}_{\text{A}}}{{\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}}$

Weight percent: The mass of one component divided by the total mass of the mixture, multiplied by $100\%$

  $\text{weight \% A = }\frac{\text{Mass of A}}{\text{ Mass of A + Mass of B + Mass of C + }\mathrm{...}}\text{×100\%}$

The number of moles of any substance can be determined using the equation

  $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent.

  $\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

Answer to Problem 56GQ

Molality of the solution is $2.67m$

Explanation of Solution

Given,

  Given mass of $NaCl$ is $1130g$

  Given mass of ice is $7250g=7.25kg$

  Molar mass of $NaCl$ is $\text{58}\text{.44}\text{g/mol}$

  Molar mass of ice is $\text{18}\text{.02}\text{g/mol}$

Number of moles of $NaCl$ is,

  $\begin{array}{c}{\text{n}}_{NaCl}\text{=}\frac{\text{1130}\text{g}}{\text{58}\text{.44}\text{g/mol}}\\ =19.336mol\end{array}$

Number of moles of ice is,

  $\begin{array}{c}{\text{n}}_{ice}\text{=}\frac{\text{7250}\text{g}}{\text{18}\text{.02}\text{g/mol}}\\ =402.33mol\end{array}$

Molality of the solution is calculated,

$\begin{array}{c}\text{Molality (m) =}\frac{19.336mol}{\text{7}\text{.25kg}}\\ =2.67m\end{array}$

Molality of the solution is $2.67m$