Chapter 3, Problem 1ASA

Using Figure 3.1, determine

a. the number of grams of ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$ that will dissolve in $100\text{ g}$ of ${\text{H}}_{\text{2}}\text{O}$ at $100°\text{C}$.

$\_\_\_\_\_\_\_\_\_ \text{g}{\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$

b. the number of grams of water required to dissolve $4.\text{0 g}$ of ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$ at $100°\text{C}$. (Hint: your answer to Problem 1(a) gives you the needed conversion factor for g ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$ to ${\text{g H}}_{\text{2}}\text{O}$.)

$\_\_\_\_\_\_\_\_\_ \text{g}{\text{H}}_{\text{2}}\text{O}$

c. the number of grams of water required to dissolve ${\text{18 g KNO}}_{\text{3}}$ at $100°\text{C}$.

$\_\_\_\_\_\_\_\_\_ \text{g}{\text{H}}_{\text{2}}\text{O}$

d. the number of grams of water required to dissolve a mixture containing ${\text{18 g KNO}}_{\text{3}}$ and $4.\text{0 g}$ ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$, assuming that the solubility of one substance is not affected by the presence of another.

$\_\_\_\_\_\_\_\_\_ \text{g}{\text{H}}_{\text{2}}\text{O}$

Interpretation Introduction

(a)

Interpretation:

The number of grams of ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$ that dissolves in $100\text{ g}$ of ${\text{H}}_{\text{2}}\text{O}$ at $100°\text{C}$ is to be stated.

Concept introduction:

Solution is a homogenous mixture of two or more components. A sample taken from any part of the solution will have the same composition as the rest of the solution. Many chemical reactions occur in water solutions.

Answer to Problem 1ASA

The number of grams of ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$ that dissolves in $100\text{ g}$ of ${\text{H}}_{\text{2}}\text{O}$ at $100°\text{C}$ is about $\text{180}\text{g}$.

Explanation of Solution

The given illustration of the graph is shown below.

Figure 1

According to the above graph, the number of grams of ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$ that dissolves in $100\text{ g}$ of ${\text{H}}_{\text{2}}\text{O}$ at $100°\text{C}$ is about $\text{180}\text{g}$.

Conclusion

The number of grams of ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$ that dissolves in $100\text{ g}$ of ${\text{H}}_{\text{2}}\text{O}$ at $100°\text{C}$ is about $\text{180}\text{g}$.

Interpretation Introduction

(b)

Interpretation:

The number of grams of water required to dissolve $4.\text{0 g}$ of ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$ at $100°\text{C}$ is to be stated.

Concept introduction:

Solution is a homogenous mixture of two or more components. A sample taken from any part of the solution will have the same composition as the rest of the solution. Many chemical reactions occur in water solutions.

Answer to Problem 1ASA

The number of grams of water required to dissolve $4.\text{0 g}$ of ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$ at $100°\text{C}$ is $2.2\text{g}{\text{H}}_{\text{2}}\text{O}$.

Explanation of Solution

The given illustration of the graph is shown below.

Figure 1

According to the above graph, the number of grams of ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$ that dissolves in $100\text{ g}$ of ${\text{H}}_{\text{2}}\text{O}$ at $100°\text{C}$ is about $\text{200}\text{g}$. So, to dissolve $\text{1}\text{g}$ of ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$ in $100\text{ g}$ of ${\text{H}}_{\text{2}}\text{O}$, the expression required is $1.\text{0 g}\times \frac{100\text{g}{\text{H}}_{\text{2}}\text{O}}{180\text{g}{\text{CuSO}}_{\text{4}}}$.

Thus, water required to dissolve $4.\text{0 g}$ of ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$ is $4.\text{0 g}\times \frac{100\text{g}{\text{H}}_{\text{2}}\text{O}}{180\text{g}{\text{CuSO}}_{\text{4}}}=2.2\text{g}{\text{H}}_{\text{2}}\text{O}$.

Conclusion

The number of grams of water required to dissolve $4.\text{0 g}$ of ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$ at $100°\text{C}$ is $2.2\text{g}{\text{H}}_{\text{2}}\text{O}$.

Interpretation Introduction

(c)

Interpretation:

The number of grams of water required to dissolve ${\text{18 g KNO}}_{\text{3}}$ at $100°\text{C}$ is to be stated.

Concept introduction:

Solution is a homogenous mixture of two or more components. A sample taken from any part of the solution will have the same composition as the rest of the solution. Many chemical reactions occur in water solutions.

Answer to Problem 1ASA

The number of grams of water required to dissolve ${\text{18 g KNO}}_{\text{3}}$ at $100°\text{C}$ is $6.67\text{g}{\text{H}}_{\text{2}}\text{O}$.

Explanation of Solution

The given illustration of the graph is shown below.

Figure 1

According to the above graph, the number of grams of ${\text{KNO}}_{\text{3}}$ that dissolves in $100\text{ g}$ of ${\text{H}}_{\text{2}}\text{O}$ at $100°\text{C}$ is about $\text{270}\text{g}$. So, to dissolve $\text{1}\text{g}$ of ${\text{KNO}}_{\text{3}}$ in $100\text{ g}$ of ${\text{H}}_{\text{2}}\text{O}$, the expression required is $1.\text{0 g}\times \frac{100\text{g}{\text{H}}_{\text{2}}\text{O}}{\text{270}\text{g}{\text{KNO}}_{\text{3}}}$.

Thus, the water required to dissolve $18.\text{0 g}$ of ${\text{KNO}}_{\text{3}}$ is $18.\text{0 g}\times \frac{100\text{g}{\text{H}}_{\text{2}}\text{O}}{270\text{g}{\text{KNO}}_{\text{3}}}=6.67\text{g}{\text{H}}_{\text{2}}\text{O}$.

Conclusion

The number of grams of water required to dissolve ${\text{18 g KNO}}_{\text{3}}$ at $100°\text{C}$ is $6.67\text{g}{\text{H}}_{\text{2}}\text{O}$.

Interpretation Introduction

(d)

Interpretation:

The number of grams of water required to dissolve a mixture containing ${\text{18 g KNO}}_{\text{3}}$ and $4.\text{0 g}$ ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$ is to be stated.

Concept introduction:

Solution is a homogenous mixture of two or more components. A sample taken from any part of the solution will have the same composition as the rest of the solution. Many chemical reactions occur in water solutions.

Answer to Problem 1ASA

The number of grams of water required to dissolve a mixture containing ${\text{18 g KNO}}_{\text{3}}$ and $4.\text{0 g}$ ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$ is $\text{8}\text{.87}\text{g}$.

Explanation of Solution

The given illustration of the graph is shown below.

Figure 1

The amount of water required to dissolve $18.\text{0 g}$ of ${\text{KNO}}_{\text{3}}$ is $6.67\text{g}{\text{H}}_{\text{2}}\text{O}$.

The amount of water required to dissolve $4.\text{0 g}$ of ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$ is $2.2\text{g}{\text{H}}_{\text{2}}\text{O}$.

Therefore, the total amount of water required to dissolve a mixture containing ${\text{18 g KNO}}_{\text{3}}$ and $4.\text{0 g}$ ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$ is $6.67\text{g}+2.2\text{g}=\text{8}\text{.87}\text{g}$

Conclusion

The number of grams of water required to dissolve a mixture containing ${\text{18 g KNO}}_{\text{3}}$ and $4.\text{0 g}$ ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$ is $\text{8}\text{.87}\text{g}$.