Chapter 16, Problem 53A

Interpretation Introduction

Interpretation: The mass of ${\text{ AgNO}}_{\text{3}}$ is to be calculated which can be dissolved in $\text{250}\text{g}$ of water at $\text{20}°\text{C}$ .

Concept Introduction: The mass of the solute dissolved in $\text{100}\text{g}$ of water at the specified temperature is a standard definition of the solute's solubility in water. Under the same temperature, solubility is constant and unaffected by the volume of the solution.

The formula for calculation of the mass ${\text{ AgNO}}_{\text{3}}$ dissolved in the solvent is using the unitary method:

  $\frac{{\text{ AgNO}}_{\text{3}}}{{\text{H}}_{\text{2}}\text{O}}\times 250\text{g}{\text{H}}_{\text{2}}\text{O}$

Answer to Problem 53A

The $\text{250}\text{g}$ of water at $\text{20}°\text{C}$ can dissolve $\text{555}\text{.0}\text{g}$ of ${\text{ AgNO}}_{\text{3}}$ .

Explanation of Solution

The data in the table indicates that $\text{222}\text{.0}\text{g}$ of ${\text{ AgNO}}_{\text{3}}$ can be dissolved in $\text{100}\text{g}$ of water.

So, $\text{100}\text{g}$ of water can dissolve $\text{222}\text{.0}\text{g}$ of ${\text{ AgNO}}_{\text{3}}$ .

Then, $\text{250}\text{g}$ of water can dissolve:

  $\frac{\text{ 222}\text{.0}\text{g}{\text{AgNO}}_{\text{3}}}{\text{100}\text{g}{\text{H}}_{\text{2}}\text{O}}\times 250\text{g}{\text{H}}_{\text{2}}\text{O}=555.\text{0}\text{g}{\text{AgNO}}_{\text{3}}$

Conclusion

Using the solubility table, it can be calculated that $\text{250}\text{g}$ of water at $\text{20}°\text{C}$ can dissolve $\text{555}\text{.0}\text{g}$ of ${\text{ AgNO}}_{\text{3}}$ .