Chapter 5, Problem 5D.16E

(a)

Interpretation Introduction

Interpretation:

When $4.0g$ of $KCl$ is dissolved in $100g$ of water, the temperature change has to be calculated.

Concept introduction:

The heat capacity at constant pressure can be expressed as follows,

    $C_P=\frac{\Delta H}{\Delta T}$

Explanation of Solution

Given,

    $4.0g$ of $KCl$ is dissolved in $100g$ of water

The temperature change is calculated as follows,

Molar enthalpies of solution $\Delta {\text{H}}_{\text{sol}}^o\text{ = 17}\text{.2 kJ }\cdot {\text{mol}}^{\text{-1}}$

  $\begin{array}{l}Thetemperaturechange(\Delta T)\text{=}\frac{\frac{\text{ 17}\text{.2 kJ }\cdot {\text{mol}}^{\text{-1}}}{74.56g\cdot {\text{mol}}^{\text{-1}}}\times 4g}{\text{4}\text{.18 J}\cdot {\text{K}}^{-1}\cdot {\text{g}}^{\text{-1}}\times 100g}\\ \\ Thetemperaturechange(\Delta T)\text{=}\frac{922.75J}{\text{418 J}\cdot {\text{K}}^{-1}}\\ \\ Thetemperaturechange(\Delta T)\text{=}2.21K\end{array}$

When $4.0g$ of $KCl$ is dissolved in $100g$ of water, the temperature change is $2.21K$.

(b)

Interpretation Introduction

Interpretation:

When $4.0g$ of $MgB{r}_2$ is dissolved in $100g$ of water, the temperature change has to be calculated.

Concept introduction:

Refer to part (a)

Explanation of Solution

Given,

    $4.0g$ of $MgB{r}_2$ is dissolved in $100g$ of water

The temperature change is calculated as follows,

Molar enthalpies of solution $\Delta {\text{H}}_{\text{sol}}^o\text{ = -185}\text{.6 kJ }\cdot {\text{mol}}^{\text{-1}}$

  $\begin{array}{l}Thetemperaturechange(\Delta T)\text{=}\frac{\frac{\text{ -185}\text{.6 kJ }\cdot {\text{mol}}^{\text{-1}}}{184.113g\cdot {\text{mol}}^{\text{-1}}}\times 4g}{\text{4}\text{.18 J}\cdot {\text{K}}^{-1}\cdot {\text{g}}^{\text{-1}}\times 100g}\\ \\ Thetemperaturechange(\Delta T)\text{=}\frac{-4032.3J}{\text{418 J}\cdot {\text{K}}^{-1}}\\ \\ Thetemperaturechange(\Delta T)\text{=}-9.65K\end{array}$

When $4.0g$ of $MgB{r}_2$ is dissolved in $100g$ of water, the temperature change is $-9.65K$.

(c)

Interpretation Introduction

Interpretation:

When $4.0g$ of $KN{O}_3$ is dissolved in $100g$ of water, the temperature change has to be calculated.

Concept introduction:

Refer to part (a)

Explanation of Solution

Given,

    $4.0g$ of $KN{O}_3$ is dissolved in $100g$ of water

The temperature change is calculated as follows,

Molar enthalpies of solution $\Delta {\text{H}}_{\text{sol}}^o\text{ = -23}\text{.8 kJ }\cdot {\text{mol}}^{\text{-1}}$

  $\begin{array}{l}Thetemperaturechange(\Delta T)\text{=}\frac{\frac{\text{ -23}\text{.8 kJ }\cdot {\text{mol}}^{\text{-1}}}{101.103g\cdot {\text{mol}}^{\text{-1}}}\times 4g}{\text{4}\text{.18 J}\cdot {\text{K}}^{-1}\cdot {\text{g}}^{\text{-1}}\times 100g}\\ \\ Thetemperaturechange(\Delta T)\text{=}\frac{-941.61J}{\text{418 J}\cdot {\text{K}}^{-1}}\\ \\ Thetemperaturechange(\Delta T)\text{=}-2.25K\end{array}$

When $4.0g$ of $KN{O}_3$ is dissolved in $100g$ of water, the temperature change is $-2.25K$.

(d)

Interpretation Introduction

Interpretation:

When $4.0g$ of $NaOH$ is dissolved in $100g$ of water, the temperature change has to be calculated.

Concept introduction:

Refer to part (a)

Explanation of Solution

Given,

    $4.0g$ of $NaOH$ is dissolved in $100g$ of water

The temperature change is calculated as follows,

Molar enthalpies of solution $\Delta {\text{H}}_{\text{sol}}^o\text{ = -44}\text{.5 kJ }\cdot {\text{mol}}^{\text{-1}}$

  $\begin{array}{l}Thetemperaturechange(\Delta T)\text{=}\frac{\frac{\text{ -44}\text{.5 kJ }\cdot {\text{mol}}^{\text{-1}}}{40g\cdot {\text{mol}}^{\text{-1}}}\times 4g}{\text{4}\text{.18 J}\cdot {\text{K}}^{-1}\cdot {\text{g}}^{\text{-1}}\times 100g}\\ \\ Thetemperaturechange(\Delta T)\text{=}\frac{-4450J}{\text{418 J}\cdot {\text{K}}^{-1}}\\ \\ Thetemperaturechange(\Delta T)\text{=}-10.65K\end{array}$

When $4.0g$ of $NaOH$ is dissolved in $100g$ of water, the temperature change is $-10.65K$.