Chapter 8, Problem 80P

Interpretation Introduction

(a)

Interpretation:

The freezing point of solution that is formed upon an addition of $\text{3 mol}$ offructose 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:

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

The formula to calculate the molarity is as follows:

  $\text{Molality}\left(\text{mol/kg}\right)\text{=}\frac{\text{Number of moles}}{\text{Mass of solvent}\left(\text{kg}\right)}$

Answer to Problem 80P

The freezing point of solution that is formed upon addition of $\text{3 mol}$ offructose in $1\text{ kg}$ of water is $-5.58°C$.

Explanation of Solution

Since fructose is non-electrolyte, each mole of fructose gives one particle. This can be represented as the conversion factor as given below:

  $\text{Number of particles per mole of sucrose}=\frac{1\text{ mol particle}}{\text{mol}\text{}\text{ of sucrose}}$

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

  $\Delta {\text{T}}_f\left(°C\right)={\text{K}}_f\text{m}$

Where,

• $\Delta {\text{T}}_f$denotes the depression in freezing point.
• ${\text{K}}_f$ is the freezing point depression constant
• $\text{m}$ denotes the molality.

The formula to calculate the molarity is as follows:

  $\text{Molality}\left(\text{mol/kg}\right)\text{=}\frac{\text{Number of moles}}{\text{Mass of solvent}\left(\text{kg}\right)}$

The number of moles of fructose is $\text{3 mol}$.

Mass of solvent water in kilograms is $1\text{ kg}$.

Substitute the values in above equation to calculate the molality.

  $\begin{array}{c}\text{Molality}\left(\text{mol/kg}\right)\text{=}\frac{\text{3 mol}}{1\text{ kg}}\\ =3\text{ mol/kg}\end{array}$

The freezing point depression constant is $1.86°\text{C/}\left(\text{mol/kg}\right)$.

Molality is found as $\text{3 mol/kg}$.

Substitute the values in above equation to calculate $\Delta {\text{T}}_f$.

  $\begin{array}{c}\Delta {\text{T}}_f={\text{K}}_f\text{m}\\ =\left(\frac{1.86°\text{C}}{\text{1 mol/kg}}\right)\left(\text{3 mol/kg}\right)\\ =5.58°C\end{array}$

The freezing point temperature can be determined by the formula given as follows:

  $\text{Freezing point}\left(°\text{C}\right)={\text{T}}_0\left(°\text{C}\right)-\Delta {\text{T}}_f\left(°\text{C}\right)$

Where,

• $\Delta {\text{T}}_f$denotes the depression in freezing point.
• ${\text{T}}_0$denotes the temperature of freezing point of water.

The value of ${\text{T}}_0$ is $0°\text{C}$.

The value of $\Delta {\text{T}}_f$ is found to be $5.58°C$.

Substitute the values in above formula to calculate the melting point temperature.

  $\begin{array}{c}\text{Melting point}\left(°\text{C}\right)={\text{T}}_0\left(°\text{C}\right)-\Delta {\text{T}}_f\left(°\text{C}\right)\\ =0°\text{C}-\left(5.58°C\right)\\ =-5.58°C\end{array}$

Interpretation Introduction

(b)

Interpretation:

The freezing point of the solution that is formed upon addition of $\text{1}\text{.2 mol}$ of $\text{KI}$ 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:

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

The formula to calculate the molarity is as follows:

  $\text{Molality}\left(\text{mol/kg}\right)\text{=}\frac{\text{Number of moles}}{\text{Mass of solvent}\left(\text{kg}\right)}$

Answer to Problem 80P

The freezing point of solution that is formed upon addition of $\text{1}\text{.2 mol KI}$ in $1\text{ kg}$ of water is $-4.464°C$.

Explanation of Solution

Since $\text{KI}$ is an electrolyte, each mole of $\text{KI}$ gives 2 particles that include one ${\text{K}}^{+}$ and ${\text{I}}^{-}$.

This can be represented as the conversion factor as given below:

  $\text{Number of particles per mole of KI}=\frac{\text{2 mol particles}}{\text{mol}\text{}\text{ of KI}}$

Therefore, the extent of decrease in temperature can be calculated as follows:

  $\begin{array}{c}\Delta {\text{T}}_f\left(°C\right)=\left(\frac{1.86°C}{molparticles}\right)\left(\text{1}\text{.2 mol KI}\right)\left(\frac{\text{2 mol particles}}{\text{mol KI}\text{}}\right)\\ =4.464°C\end{array}$

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

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

Where,

• $\Delta {\text{T}}_f$denotes the depression in freezing point.
• ${\text{K}}_f$is the freezing point depression constant
• $\text{m}$ denotes the molality.
• $i$ denotes van't Hoff factor.

The formula to calculate the molarity is as follows:

  $\text{Molality}\left(\text{mol/kg}\right)\text{=}\frac{\text{Number of moles}}{\text{Mass of solvent}\left(\text{kg}\right)}$

The number of moles of $\text{KI}$ is $\text{1}\text{.2 mol}$.

Mass of solvent water in kilograms is $1\text{ kg}$.

Substitute the values in above equation to calculate the molality.

  $\begin{array}{c}\text{Molality}\left(\text{mol/kg}\right)\text{=}\frac{\text{1}\text{.2 mol}}{1\text{ kg}}\\ =1.2\text{ mol/kg}\end{array}$

The freezing point depression constant is $1.86°\text{C}/\left(\text{mol/kg}\right)$.

Molality is found as $1.2\text{ mol/kg}$.

Value of $i$ is 2.

Substitute the values in above equation to calculate $\Delta {\text{T}}_f$.

  $\begin{array}{c}\Delta {\text{T}}_f=i{\text{K}}_f\text{m}\\ =\left(2\right)\left(\frac{1.86°\text{C}}{\text{1 mol/kg}}\right)\left(1.2\text{ mol/kg}\right)\\ =4.464°C\end{array}$

The freezing point temperature can be determined by the formula given as follows:

  $\text{Freezing point}\left(°\text{C}\right)={\text{T}}_0\left(°\text{C}\right)-\Delta {\text{T}}_f\left(°\text{C}\right)$

Where,

• $\Delta {\text{T}}_f$denotes the depression in freezing point.
• ${\text{T}}_0$denotes the temperature of freezing point of water.

The value of ${\text{T}}_0$ is $0°\text{C}$.

The value of $\Delta {\text{T}}_f$ is found to be $4.464°C$.

Substitute the values in the above formula to calculate the melting point temperature.

  $\begin{array}{c}\text{Freezing point}\left(°\text{C}\right)={\text{T}}_0\left(°\text{C}\right)-\Delta {\text{T}}_f\left(°\text{C}\right)\\ =0°\text{C}-\left(4.464°C\right)\\ =-4.464°C\end{array}$

Interpretation Introduction

(c)

Interpretation:

The freezing point of solution that is formed upon addition of $\text{1}\text{.5 mol}$ of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ 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:

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

The formula to calculate the molarity is as follows:

  $\text{Molality}\left(\text{mol/kg}\right)\text{=}\frac{\text{Number of moles}}{\text{Mass of solvent}\left(\text{kg}\right)}$

Answer to Problem 80P

The freezing point of solution that is formed upon addition of $\text{1}\text{.5 mol}$ of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ in $1\text{ kg}$ of water is $-11.16°C$.

Explanation of Solution

Since ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ is an electrolyte, each mole of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ gives 4 particles. This can be represented as the conversion factor as given below:

  ${\text{Number of particles per mole of Na}}_{\text{3}}{\text{PO}}_{\text{4}}\text{= }\frac{\text{4 mol particles}}{\text{mol}\text{}{\text{ of Na}}_{\text{3}}{\text{PO}}_{\text{4}}}$

Therefore, the extent of decrease in temperature can be calculated as follows:

  $\begin{array}{c}\Delta {\text{T}}_f\left(°C\right)=\left(\frac{1.86°C}{molparticles}\right)\left(\text{1}\text{.5 mol }\right)\left(\frac{\text{4 mol particle}}{\text{mol}\text{}}\right)\\ =11.16°C\end{array}$

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

  $\Delta {\text{T}}_f\left(°C\right)={\text{K}}_f\text{m}$

Where,

• $\Delta {\text{T}}_f$ denotes the depression in freezing point.
• ${\text{K}}_f$ is the freezing point depression constant
• $\text{m}$ denotes the molality.

The formula to calculate the molarity is as follows:

  $\text{Molality}\left(\text{mol/kg}\right)\text{=}\frac{\text{Number of moles}}{\text{Mass of solvent}\left(\text{kg}\right)}$

The number of molesof ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ is $\text{1}\text{.5 mol}$.

Mass of solvent water in kilograms is $1\text{ kg}$.

Substitute the values in above equation to calculate the molality.

  $\begin{array}{c}\text{Molality}\left(\text{mol/kg}\right)\text{=}\frac{\text{1}\text{.5 mol}}{1\text{ kg}}\\ =1.5\text{ mol/kg}\end{array}$

The freezing point depression constant is $1.86°\text{C}/\left(\text{mol/kg}\right)$.

Molality is found as $\text{1}\text{.5 mol/kg}$.

Substitute the values in the above equation to calculate $\Delta {\text{T}}_f$.

  $\begin{array}{c}\Delta {\text{T}}_f\left(°C\right)=i{\text{K}}_f\text{m}\\ =\left(4\right)\left(\frac{1.86°\text{C}}{\text{1 mol/kg}}\right)\left(1.5\text{mol/kg}\right)\\ =11.16°C\end{array}$

The freezing point temperature can be determined by the formula given as follows:

  $\text{Freezing point}\left(°\text{C}\right)={\text{T}}_0\left(°\text{C}\right)-\Delta {\text{T}}_f\left(°\text{C}\right)$

Where,

• $\Delta {\text{T}}_f$denotes the depression in freezing point.
• ${\text{T}}_0$ denotes the temperature of freezing point of water.

The value of ${\text{T}}_0$ is $0°\text{C}$.

The value of $\Delta {\text{T}}_f$ is found to be $11.16°C$.

Substitute the values in the above formula to calculate the melting point temperature.

  $\begin{array}{c}\text{Freezing point}\left(°\text{C}\right)={\text{T}}_0\left(°\text{C}\right)-\Delta {\text{T}}_f\left(°\text{C}\right)\\ =0°\text{C}-\left(11.16°C\right)\\ =-11.16°C\end{array}$