Chapter 8.8, Problem 8.18PP

Interpretation Introduction

(a)

Interpretation:

The melting point of solution that is formed upon addition of $2\text{ mol}$ of sucrose in $1\text{ kg}$ of water is to be determined.

Concept introduction:

The effect of addition of a non-volatile component to a solvent is that the freezing point of the solvent is reduced. This is called freezing point depression. The extent of freezing point depression is directly proportional to the amount of non-volatile solute added.

Addition of one mole of any non-volatile solute decreases the freezing point of one kilogram of water by $1.86°\text{C}$.

For example, $\text{NaCl}$ dissociates to give ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ . Since there are two particles per mole of $\text{NaCl}$ , it reduces the freezing point calculated as follows:

  $\begin{array}{c}\text{Depression in freezing point}\left(°\text{C}\right)=2\left(-1.86°\text{C}\right)\\ =-3.72°\text{C}\end{array}$

The formula to calculate number of moles from mass is given asfollows:

  $\text{Number of moles}=\frac{\text{given mass}\left(\text{g}\right)}{\text{molar mass}\left(\text{g/mol}\right)}$

The formula to calculate the depression in freezing point is as follows:

  $\Delta {\text{T}}_f={\text{K}}_f\text{m}$

Interpretation Introduction

(b)

Interpretation:

The melting point of solution that is formed upon addition of $2\text{ mol}$ of ${\text{CaCl}}_{\text{2}}$ in $1\text{ kg}$ of water is to be determined.

Concept introduction:

The effect of addition of a non-volatile component to a solvent is that the freezing point of the solvent is reduced. This is called freezing point depression. The extent of freezing point depression is directly proportional to the amount of non-volatile solute added.

Addition of one mole of any non-volatile solute decreases the freezing point of one kilogram of water by $1.86°\text{C}$.

For example, $\text{NaCl}$ dissociates to give ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ . Since there are two particles per mole of $\text{NaCl}$ , it reduces the freezing point calculated as follows:

  $\begin{array}{c}\text{Depression in freezing point}\left(°\text{C}\right)=2\left(-1.86°\text{C}\right)\\ =-3.72°\text{C}\end{array}$

The formula to calculate number of moles from mass is given asfollows:

  $\text{Number of moles}=\frac{\text{given mass}\left(\text{g}\right)}{\text{molar mass}\left(\text{g/mol}\right)}$

The formula to calculate the depression in freezing point is as follows:

  $\Delta {\text{T}}_f={\text{K}}_f\text{m}$

Interpretation Introduction

(c)

Interpretation:

The melting point of solution that is formed upon addition of $2\text{ mol}$ of ${\text{KNO}}_{\text{3}}$ in $1\text{ kg}$ of water is to be determined.

Concept introduction:

The effect of addition of a non-volatile component to a solvent is that the freezing point of the solvent is reduced. This is called freezing point depression. The extent of freezing point depression is directly proportional to the amount of non-volatile solute added.

Addition of one mole of any non-volatile solute decreases the freezing point of one kilogram of water by $1.86°\text{C}$.

For example, $\text{NaCl}$ dissociates to give ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ . Since there are two particles per mole of $\text{NaCl}$ , it reduces the freezing point calculated as follows:

  $\begin{array}{c}\text{Depression in freezing point}\left(°\text{C}\right)=2\left(-1.86°\text{C}\right)\\ =-3.72°\text{C}\end{array}$

The formula to calculate number of moles from mass is given asfollows:

  $\text{Number of moles}=\frac{\text{given mass}\left(\text{g}\right)}{\text{molar mass}\left(\text{g/mol}\right)}$

The formula to calculate the depression in freezing point is as follows:

  $\Delta {\text{T}}_f={\text{K}}_f\text{m}$

Interpretation Introduction

(d)

Interpretation:

The melting point of solution that is formed upon addition of $20\text{ g}$ of $\text{NaCl}$ in $1\text{ kg}$ of water is to be determined.

Concept introduction:

The effect of addition of a non-volatile component to a solvent is that the freezing point of the solvent is reduced. This is called freezing point depression. The extent of freezing point depression is directly proportional to the amount of non-volatile solute added.

Addition of one mole of any non-volatile solute decreases the freezing point of one kilogram of water by $1.86°\text{C}$.

For example, $\text{NaCl}$ dissociates to give ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ . Since there are two particles per mole of $\text{NaCl}$ , it reduces the freezing point calculated as follows:

  $\begin{array}{c}\text{Depression in freezing point}\left(°\text{C}\right)=2\left(-1.86°\text{C}\right)\\ =-3.72°\text{C}\end{array}$

The formula to calculate number of moles from mass is given asfollows:

  $\text{Number of moles}=\frac{\text{given mass}\left(\text{g}\right)}{\text{molar mass}\left(\text{g/mol}\right)}$

The formula to calculate the depression in freezing point is as follows:

  $\Delta {\text{T}}_f={\text{K}}_f\text{m}$