Chapter 8, Problem 79P

What is the boiling point of a solution that contains each of the following quantitiesof solute in 1.00 kg of water?

1. 3. 0molof fructose molecules
2. 1.2 mol of KI
3. 1.5 mol of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$

Interpretation Introduction

(a)

Interpretation:

The boiling point of solution formed upon the addition of $\text{3 mol}$ fructose to one kilogram of water is to be determined.

Concept introduction:

The effect of addition of a non-volatile component to a solvent is that there is anincreasein the boiling point. This is called elevation in boiling point. Such an increase is observed since the surface of the solvent is saturated with solute particles. This results in lesser escape of gaseous molecules and hence the vapor pressure of the solution gets lowered compared to the vapor pressure found above the pure solvent to which no solute has been added.

The temperature at which vapor pressure of solution becomes equal to the atmospheric pressure is known as the boiling point. Dueto the lesser vapor pressure, the solution needs to be heated to higher temperature in order to boil it. Therefore the boiling point of the solution is elevated.

One mole of nonvolatile solute increases the boiling point of one kilogram of water by $0.51°\text{C}$.

The extent of elevation in boiling point is directly proportional to the amount of non-volatile solute added. For example, $\text{NaCl}$ dissociates to give ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ . Since there are two particles per mole of $\text{NaCl}$ , it increases the boiling point calculated as follows:

  $\begin{array}{c}\text{Elevation in boiling point}\left(°\text{C}\right)=2\left(0.51°\text{C}\right)\\ =1.02°\text{C}\end{array}$

The formula to calculate the 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 molality 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 formula to calculate the elevation in boiling point is as follows:

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

The formula to calculate the molarity is as follows:

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

The formula to calculate $\Delta {\text{T}}_b$ is as follows:

  $\Delta {\text{T}}_b={\text{T}}_b{}^0-{\text{T}}_b$

Answer to Problem 79P

The boiling point of solution formed upon addition of $\text{3 mol}$ fructose to one kilogram of water is $101.53°C$.

Explanation of Solution

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

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

Alternatively, the formula to calculate the elevation in boiling point is as follows:

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

Where,

• $\Delta {\text{T}}_b$ denotes the elevation in boiling point.
• ${\text{K}}_b$ is the elevation in boiling point constant.
• $\text{m}$denotes the molality.

The formula to calculate the molality 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 elevation in boiling point constant is $0.51°\text{C}/\left(\text{mol/kg}\right)$.

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

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

  $\begin{array}{c}\Delta {\text{T}}_b\left(°C\right)={\text{K}}_b\text{m}\\ =\left(\frac{0.51°\text{C}}{\text{1 mol/kg}}\right)\left(\text{3 mol/kg}\right)\\ =1.53°C\end{array}$

The formula to calculate $\Delta {\text{T}}_b$ is as follows:

  $\Delta {\text{T}}_b={\text{T}}_b-{\text{T}}_b{}^0$

Where,

• ${\text{T}}_b$ denotes the temperature at which the solution boils after the nonvolatile solute has been added.
• ${\text{T}}_b{}^0$ denotes the temperature of boiling point of water.

The value of ${\text{T}}_b{}^0$ is $100°\text{C}$.

The value of $\Delta {\text{T}}_b$ is $1.53°C$.

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

  $1.53°C={\text{T}}_b-100°\text{C}$

Rearrange the above formula to calculate the boiling point of the solution.

  $\begin{array}{c}{\text{T}}_b\left(°C\right)=100°\text{C}+1.53°C\\ =101.53°C\end{array}$

Therefore the boiling point of the solution is $101.53°C$.

Interpretation Introduction

(b)

Interpretation:

The boiling point of solution formed upon addition of $\text{1}\text{.2 mol}$

  $\text{KI}$ to on kilogram of water is to be determined.

Concept introduction:

The effect of addition of a non-volatile component to a solvent is that there is anincreasein the boiling point. This is called elevation in boiling point. Such an increase is observed since the surface of the solvent is saturated with solute particles. This results in lesser escape of gaseous molecules, and hence the vapor pressure of the solution gets lowered compared to the vapor pressure found above the pure solvent to which no solute has been added.

The temperature at which vapor pressure of solution becomes equal to the atmospheric pressure is known as the boiling point. Dueto the lesser vapor pressure, the solution needs to be heated to higher temperature in order to boil it. Therefore the boiling point of the solution is elevated.

One mole of nonvolatile solute increases the boiling point of one kilogram of water by $0.51°\text{C}$.

The extent of elevation in boiling point is directly proportional to the amount of non-volatile solute added. For example, $\text{NaCl}$ dissociates to give ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ . Since there are two particles per mole of $\text{NaCl}$ , it increases the boiling point calculated as follows:

  $\begin{array}{c}\text{Elevation in boiling point}\left(°\text{C}\right)=2\left(0.51°\text{C}\right)\\ =1.02°\text{C}\end{array}$

The formula to calculate the 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 molality 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 formula to calculate the elevation in boiling point is as follows:

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

The formula to calculate the molarity is as follows:

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

The formula to calculate $\Delta {\text{T}}_b$ is as follows:

  $\Delta {\text{T}}_b={\text{T}}_b{}^0-{\text{T}}_b$

Answer to Problem 79P

The boiling point of solution formed upon addition of $\text{1}\text{.2 mol KI}$ on kilogram of water is $101.224°C$.

Explanation of Solution

$\text{KI}$ dissociates to give ${\text{K}}^{+}$ and ${\text{I}}^{-}$ . So, each mole of $\text{KI}$ gives $\text{2 mol}$ particles. This can be represented as the conversion factor as given below:

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

Alternatively, the formula to calculate the elevation in boiling point is as follows:

  $\Delta {\text{T}}_b={\text{iK}}_b\text{m}$

Where,

  $\Delta {\text{T}}_b$ denotes the elevation in boiling point.

  ${\text{K}}_b$ is the freezing point depression constant.

  $\text{m}$ denotes the molality.

  $\text{i}$ denotes the Van't Hoff factor.

Van't Hoff factor refers to the number of particles into which the salt is dissociated.

Since $\text{KI}$ dissociates to give ${\text{K}}^{+}$ and ${\text{I}}^{-}$ .Therefore two particles are produced and so the value of $\text{i}$ is 2.

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 the 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 elevation in boiling point constant is $0.51°\text{C}/\left(\text{mol/kg}\right)$.

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

The value of $\text{i}$ is 2.

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

  $\begin{array}{c}\Delta {\text{T}}_b\left(°C\right)={\text{iK}}_b\text{m}\\ =\left(2\right)\left(\frac{0.51°\text{C}}{\text{1 }\cancel{\text{mol/kg}}}\right)\left(\text{1}\text{.2 }\cancel{\text{mol/kg}}\right)\\ =1.224°C\end{array}$

The formula to calculate $\Delta {\text{T}}_b$ is as follows:

  $\Delta {\text{T}}_b={\text{T}}_b-{\text{T}}_b{}^0$

Where,

• ${\text{T}}_b$ denotes the temperature at which the solution boils after the nonvolatile solute has been added.
• ${\text{T}}_b{}^0$ denotes the temperature of boiling point of water.

The value of ${\text{T}}_b{}^0$ is $100°\text{C}$.

The value of $\Delta {\text{T}}_b$ is $1.224°C$.

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

  $1.224°C={\text{T}}_b-100°\text{C}$

Rearrange the above formula to calculate the boiling point of the solution.

  $\begin{array}{c}{\text{T}}_b\left(°C\right)=100°\text{C}+1.224°C\\ =101.224°C\end{array}$

Interpretation Introduction

(c)

Interpretation:

The boiling point of solution formed upon addition of $\text{1}{\text{.5 mol Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ in one kilogram of water is to be determined.

Concept introduction:

The effect of addition of a non-volatile component to a solvent is that there is anincreasein the boiling point. This is called elevation in boiling point. Such an increase is observed since the surface of the solvent is saturated with solute particles. This results in lesser escape of gaseous molecules, and hence the vapor pressure of the solution gets lowered compared to the vapor pressure found above the pure solvent to which no solute has been added.

The temperature at which vapor pressure of solution becomes equal to the atmospheric pressure is known as the boiling point. Dueto the lesser vapor pressure, the solution needs to be heated to higher temperature in order to boil it. Therefore the boiling point of the solution is elevated.

One mole of nonvolatile solute increases the boiling point of one kilogram of water by $0.51°\text{C}$.

The extent of elevation in boiling point is directly proportional to the amount of non-volatile solute added. For example, $\text{NaCl}$ dissociates to give ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ . Since there are two particles per mole of $\text{NaCl}$ , it increases the boiling point calculated as follows:

  $\begin{array}{c}\text{Elevation in boiling point}\left(°\text{C}\right)=2\left(0.51°\text{C}\right)\\ =1.02°\text{C}\end{array}$

The formula used for the calculation of number of moles from mass is 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 molality 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 formula to calculate the elevation in boiling point is as follows:

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

The formula to calculate the molarity is as follows:

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

The formula to calculate $\Delta {\text{T}}_b$ is as follows:

  $\Delta {\text{T}}_b={\text{T}}_b{}^0-{\text{T}}_b$

Answer to Problem 79P

The boiling point of solution formed upon addition of $\text{1}{\text{.5 mol Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ in one kilogram of water is

Explanation of Solution

${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ dissociates to give ${\text{3Na}}^{+}$ and ${\text{PO}}_{\text{4}}{}^{3-}$ . So, each mole of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ gives three particles. This can be represented as the conversion factor as given below:

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

Alternatively, the formula to calculate theelevation in boiling point is as follows:

  $\Delta {\text{T}}_b={\text{iK}}_b\text{m}$

Where,

• $\Delta {\text{T}}_b$ denotes the elevation in boiling point.
• ${\text{K}}_b$ is the elevation in boiling point constant.
• $\text{m}$denotes the molality.
• $\text{i}$ denotes the Van't Hoff factor.

Van't Hoff factor refers to the number of particles into which the salt is dissociated.

Since ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ dissociates to give ${\text{3Na}}^{+}$ and ${\text{PO}}_{\text{4}}{}^{3-}$ therefore two particles are produced and so the value of $\text{i}$ is 4.

The formula to calculate the molality 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 ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ is $\text{1}\text{.5 mol}$.

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

Substitute the values in the 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 elevation in boiling point constant is $0.51°\text{C}/\left(\text{mol/kg}\right)$.

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

The value of $\text{i}$ is 4.

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

  $\begin{array}{c}\Delta {\text{T}}_b\left(°C\right)={\text{iK}}_b\text{m}\\ =\left(4\right)\left(\frac{0.51°\text{C}}{\text{1 mol/kg}}\right)\left(\text{1}\text{.5 mol/kg}\right)\\ =3.06°C\end{array}$

The formula to calculate $\Delta {\text{T}}_b$ is as follows:

  $\Delta {\text{T}}_b={\text{T}}_b-{\text{T}}_b{}^0$

Where,

• ${\text{T}}_b$ denotes the temperature at which the solution boils after the nonvolatile solute has been added.
• ${\text{T}}_b{}^0$ denotes the temperature of boiling point of water.

The value of ${\text{T}}_b{}^0$ is $100°\text{C}$.

The value of $\Delta {\text{T}}_b$ is $3.06°C$.

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

  $3.06°C={\text{T}}_b-100°\text{C}$

Rearrange the above formula to calculate the boiling point of the solution.

  $\begin{array}{c}{\text{T}}_b\left(°C\right)=100°\text{C}+3.06°C\\ =103.06°C\end{array}$