Chapter 16, Problem 95A

Interpretation Introduction

Interpretation: The solution with greater elevation in boiling point is to be calculated that has the solute, and calcium nitrate in different masses.

Concept introduction: The addition of solute particles to a particular solvent will decrease the freezing point and raise the boiling point of that solvent.

The formula used for calculating the elevation in boiling point is,

  $\Delta T_b=K_{b}m$

Here $\Delta T_b$ is the elevation in boiling point, $K_b$ is the molal boiling point elevation constant, and $m$ is molality.

The expression for moles is as follows:

  $\text{moles}=\frac{\text{ mass of solute}}{\text{Molar mass of solute}}$   ...... (1)

The expression for molality is as follows:

  $m=\frac{\text{moles of solute}}{\text{Mass of solvent(kg)}}$   ...... (2)

Answer to Problem 95A

The elevation in the boiling point of calcium nitrate is higher value in the case of its mass being $6\text{g}$ and the mass of the solvent being $30\text{g}$ .

Explanation of Solution

The one with a greater elevation in boiling point can be calculated by calculating the elevation in the boiling point of calcium nitrate for each mass.

The given mass of solute, $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2$ is $3\text{g}$ .

The molecular mass of ( $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2$ ) is 164.

Substitute, $3\text{g}$ for the mass of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2$ and $164\text{g/mol}$ for molar mass of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2$ in equation (1).

  $\begin{array}{c}\text{moles}\left(\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2\right)=\frac{\text{ 3 g}}{\text{164 g/mol}}\\ =0.018\text{ mol}\end{array}$

Mass of the solvent, water is $60\text{g}$ .

Convert $\text{g}$ to $\text{kg}$ .

  $\begin{array}{c}60.0\text{g}=60\times \frac{1\text{kg}}{1000\text{g}}\\ =0.06\text{kg}\end{array}$

Substitute $0.018moles$ for $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2$ and $\text{0}\text{.06}\text{kg}$ for the mass of solvent in equation (2).

  $\begin{array}{c}m=\frac{0.018}{0.06}\\ =0.3\text{m}\end{array}$

The molal freezing point depression constant of water is $0.52°{\text{Cmol}}^{-1}$ .

So,

Substitute $0.52°{\text{Cmol}}^{-1}$ for $K_f$ and $0.3\text{m}$ for $m$ in the equation of freezing point depression.

  $\begin{array}{c}\Delta T_f=0.52°{\text{Cmol}}^{-1}\times \text{ 0}\text{.3}\text{mol/kg}\\ =0.156°\text{C}\end{array}$

The given mass of solute, $\text{Ca}{\left({\text{NO}}_{\text{3;}}\right)}_2$ is $6\text{g}$ .

The molecular mass of ( $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2$ ) is 164.

Substitute, $6\text{g}$ for mass of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2$ and $164\text{g/mol}$ for molar mass of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2$ in equation (1).

  $\begin{array}{c}\text{moles}\left(\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2\right)=\frac{\text{ 6 g}}{\text{164 g/mol}}\\ =0.036\text{ mol}\end{array}$

Mass of the solvent, water is $30\text{g}$ .

Convert $\text{g}$ to $\text{kg}$ .

  $\begin{array}{c}30.0\text{g}=30\times \frac{1\text{kg}}{1000\text{g}}\\ =0.03\text{kg}\end{array}$

Substitute $0.036\text{ mol}$ for $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2$ and $\text{0}\text{.03}\text{kg}$ for the mass of solvent in equation (2).

  $\begin{array}{c}m=\frac{0.036}{0.03}\\ =1.2\text{m}\end{array}$

The molal freezing point depression constant of water is $0.52°{\text{Cmol}}^{-1}$ .

So,

Substitute $0.52°{\text{Cmol}}^{-1}$ for $K_f$ and $1.2\text{m}$ for $m$ in the equation of freezing point depression.

  $\begin{array}{c}\Delta T_f=0.52°{\text{Cmol}}^{-1}\times \text{ 1}\text{.2}\text{mol/kg}\\ =0.624°\text{C}\end{array}$

The elevation in boiling point, ΔTb is higher for calcium nitrate with a mass of 6g and a mass of solvent 30g.

Conclusion

Hence, the elevation in the boiling point of calcium nitrate is higher value in the case of its mass being $6\text{g}$ and the mass of the solvent being $30\text{g}$ .