Chapter 13.3, Problem 13.2WE

(a)

Interpretation Introduction

Interpretation:

The concentration of a solution has to be expressed in terms of molality, percent by mass and parts per million.

Concept introduction:

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent. Molality is estimation of moles in relationship with solvent in the solution.

$\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

Percent by mass: Mass percent is mass of the element is divided by total mass of the compound and multiplied by 100.

$\text{Percent mass =}\frac{\text{Mass}\text{of}\text{the}\text{element}}{\text{total}\text{mass}\text{of}\text{the}\text{compound}}\text{×100\%}$

Answer to Problem 13.2WE

The concentration of the solution in terms of molality (m) = $\text{1}\text{.06m}$

Explanation of Solution

Calculation of number of moles of glucose solution

$\frac{\text{170}\text{.1g}}{\text{180}\text{.2g/mol}}\text{= 0}\text{.9440 mol}$

By plugging in the value of mass and molar mass of glucose, the number of moles of the glucose solution has calculated.

Calculation of mass of litre of solution

$\text{1liter solution ×}\frac{\text{1062g}}{\text{L}}\text{=1062g}$

By multiplying in the value of density of the solution per liter, mass of litre of solution has calculated.

Calculation of molality of glucose solution

$\text{1062 g solution -170}\text{.1g glucose = 892g water = 0}\text{.892kg water}$

$\frac{\text{0}\text{.9440mol}}{\text{0}\text{.892kg}}\text{= 1}\text{.06m}$

By plugging in the value of moles of the glucose and kilogram of water, the molality solution has calculated.

Conclusion

The concentration of the solution in terms of molality (m) has calculated as $\text{1}\text{.06m}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The concentration of a solution has to be expressed in terms of molality, percent by mass and parts per million.

Concept introduction:

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent. Molality is estimation of moles in relationship with solvent in the solution.

$\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

Percent by mass: Mass percent is mass of the element is divided by total mass of the compound and multiplied by 100.

$\text{Percent mass =}\frac{\text{Mass}\text{of}\text{the}\text{element}}{\text{total}\text{mass}\text{of}\text{the}\text{compound}}\text{×100\%}$

Answer to Problem 13.2WE

The concentration of the solution in terms of percent by mass = $\text{16}\text{.02\%}$

Explanation of Solution

Calculation of mass percent of glucose solution

$\frac{\text{170}\text{.1g glucose}}{\text{1062 g solution}}\text{×100\% =16}\text{.02\%}$

By plugging in the value of mass of glucose, mass of water and they are multiplied by 100, mass percent of glucose solution has calculated.

Conclusion

The concentration of the solution in terms of percent by mass has calculated as $\text{16}\text{.02\%}$

(c)

Interpretation Introduction

Interpretation:

The concentration of a solution has to be expressed in terms of molality, percent by mass and parts per million.

Concept introduction:

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent. Molality is estimation of moles in relationship with solvent in the solution.

$\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

Percent by mass: Mass percent is mass of the element is divided by total mass of the compound and multiplied by 100.

$\text{Percent mass =}\frac{\text{Mass}\text{of}\text{the}\text{element}}{\text{total}\text{mass}\text{of}\text{the}\text{compound}}\text{×100\%}$

Answer to Problem 13.2WE

The concentration of the solution in terms of parts per million= $\text{1602}\times {\text{10}}^{\text{5}}\text{ppm}$

Explanation of Solution

Calculation of mass parts per million of glucose solution

$\frac{\text{170}\text{.1g glucose}}{\text{1062 g solution}}\text{×1,000,000 =1}{\text{.602×10}}^{\text{5}}\text{ppm}$

By plugging in the value of mass of glucose, mass of water and they are multiplied by million, parts per million of glucose solution has calculated.

Conclusion:

The concentration of the solution in terms of parts per million has calculated as  $\text{1602}\times {\text{10}}^{\text{5}}\text{ppm}$